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301. What is entropy a measure of?
ⓐ. Energy stored in chemical bonds
ⓑ. Degree of disorder or randomness of a system
ⓒ. Rate of a chemical reaction
ⓓ. Enthalpy change during a process
Correct Answer: Degree of disorder or randomness of a system
Explanation: Entropy (S) quantifies the randomness or disorganization of particles in a system. Higher entropy means greater disorder and more possible microscopic arrangements (microstates). For example, gases have higher entropy than liquids, and liquids have higher entropy than solids due to increased particle movement.
302. Which of the following states of matter has the highest entropy?
ⓐ. Solid
ⓑ. Liquid
ⓒ. Plasma
ⓓ. Gas
Correct Answer: Gas
Explanation: In gases, particles move freely and occupy a larger volume, leading to maximum randomness. Hence, gases have the highest entropy, followed by liquids and then solids. The more ways particles can arrange and move, the greater the entropy of the system.
303. What happens to entropy when ice melts into water at 0°C?
ⓐ. It decreases
ⓑ. It remains constant
ⓒ. It increases
ⓓ. It becomes zero
Correct Answer: It increases
Explanation: When ice melts, its molecules move from a highly ordered solid state to a less ordered liquid state. This increase in molecular freedom and number of possible arrangements raises entropy. Therefore, ΔS is positive for melting, reflecting increased disorder.
304. Which of the following represents a decrease in entropy?
ⓐ. Dissolution of salt in water
ⓑ. Evaporation of alcohol
ⓒ. Combustion of methane
ⓓ. Freezing of water
Correct Answer: Freezing of water
Explanation: Freezing converts a disordered liquid into a highly ordered crystalline solid. The number of possible molecular arrangements decreases, reducing entropy. Hence, ΔS is negative for freezing or condensation processes where order increases.
305. The change in entropy \(\Delta S\) for a reversible process is given by:
ⓐ. \(\Delta S = \dfrac{q_{\mathrm{rev}}}{T}\)
ⓑ. \(\Delta S = qT\)
ⓒ. \(\Delta S = q \times T^2\)
ⓓ. \(\Delta S = \dfrac{T}{q}\)
Correct Answer: \(\Delta S = \dfrac{q_{\mathrm{rev}}}{T}\)
Explanation: By definition, entropy change for a reversible path is \(\Delta S = \displaystyle\int \frac{\delta q_{\mathrm{rev}}}{T}\). For an isothermal reversible step this reduces to \(\Delta S = \frac{q_{\mathrm{rev}}}{T}\). The subscript “rev” is essential; using actual heat ( q ) from an irreversible path would not give the correct entropy change. Dimensions are J K\(^{-1}\), consistent with energy divided by absolute temperature.
306. Which of the following has the greatest increase in entropy?
ⓐ. Freezing of water
ⓑ. Condensation of steam
ⓒ. Sublimation of iodine
ⓓ. Crystallization of salt
Correct Answer: Sublimation of iodine
Explanation: Sublimation converts a solid directly into vapor, resulting in a huge increase in molecular randomness and freedom of motion. This transition involves the largest positive ΔS because the gaseous state has much higher entropy than either solid or liquid states.
307. In which of the following cases is entropy expected to decrease?
ⓐ. Expansion of a gas
ⓑ. Dissolution of a solid in water
ⓒ. Compression of a gas
ⓓ. Melting of ice
Correct Answer: Compression of a gas
Explanation: Compression forces gas molecules into a smaller volume, reducing the number of accessible microstates and limiting molecular motion. This decreases the disorder of the system, leading to a negative entropy change.
308. Which of the following systems has zero entropy at 0 K according to the third law of thermodynamics?
ⓐ. A gas at 0 K
ⓑ. A perfect crystalline solid at 0 K
ⓒ. A liquid at 0 K
ⓓ. A mixture of two solids at 0 K
Correct Answer: A perfect crystalline solid at 0 K
Explanation: The third law of thermodynamics states that the entropy of a perfectly ordered crystalline solid approaches zero as temperature approaches absolute zero (0 K). At this point, only one microstate exists, corresponding to perfect order and zero randomness.
309. Which of the following changes leads to an increase in entropy?
ⓐ. Dissociation of \(N_2O_4(g\) into \(2NO_2(g\)
ⓑ. Compression of gas
ⓒ. Freezing of water
ⓓ. Deposition of iodine
Correct Answer: Dissociation of \(N_2O_4(g\) into \(2NO_2(g\)
Explanation: The decomposition of dinitrogen tetroxide increases the number of gas molecules, thereby increasing randomness. More particles mean more possible microstates and thus higher entropy. Hence, ΔS is positive for this reaction.
310. Which of the following statements correctly describes entropy?
ⓐ. Entropy measures the heat content of a system
ⓑ. Entropy decreases with temperature increase
ⓒ. Entropy is a state function representing disorder
ⓓ. Entropy depends on the reaction path
Correct Answer: Entropy is a state function representing disorder
Explanation: Entropy depends only on the initial and final states of the system, not on the process path. It quantifies the degree of molecular disorder or the number of microstates accessible to a system. Since it is a state function, its change depends solely on the system’s starting and ending conditions.
311. What is the expression for entropy change (ΔS) in a reversible process?
ⓐ. \(\Delta S = q_{rev} \times T\)
ⓑ. \(\Delta S = \dfrac{q_{rev}}{T}\)
ⓒ. \(\Delta S = T \times q_{rev}\)
ⓓ. \(\Delta S = \dfrac{T}{q_{rev}}\)
Correct Answer: \(\Delta S = \dfrac{q_{rev}}{T}\)
Explanation: For a reversible process, the entropy change is the heat absorbed reversibly divided by the absolute temperature. This relationship assumes equilibrium between system and surroundings at every step, so the system remains quasi-static. It is the fundamental thermodynamic definition of entropy change.
312. What is the entropy change of the universe \(ΔS_univ\) for a reversible process?
ⓐ. Positive
ⓑ. Negative
ⓒ. Zero
ⓓ. Infinite
Correct Answer: Zero
Explanation: In a reversible process, the system and surroundings change in such a way that their entropy changes exactly cancel. Therefore, the total entropy of the universe remains constant. No net increase or decrease in disorder occurs, making ΔS_univ = 0.
313. What is the entropy change of the universe for an irreversible process?
ⓐ. Zero
ⓑ. Negative
ⓒ. Positive
ⓓ. Cannot be determined
Correct Answer: Positive
Explanation: Irreversible processes always increase the total entropy of the universe. While the system’s entropy may decrease locally, the surroundings’ entropy increases more than enough to make the total change positive. This reflects the natural tendency toward greater disorder and is a fundamental statement of the second law of thermodynamics.
314. Which of the following statements about entropy in reversible and irreversible processes is correct?
ⓐ. Entropy of the universe decreases in both processes
ⓑ. Entropy of the universe increases in reversible processes
ⓒ. Entropy of the universe increases in irreversible processes and remains constant in reversible ones
ⓓ. Entropy of the universe is zero in all cases
Correct Answer: Entropy of the universe increases in irreversible processes and remains constant in reversible ones
Explanation: The second law states that ΔS_universe ≥ 0. For reversible processes, ΔS_universe = 0, and for irreversible processes, ΔS_universe > 0. Thus, only reversible changes are thermodynamically ideal, while real natural processes are irreversible and lead to increasing entropy.
315. The melting of ice at 0°C and 1 atm is an example of:
ⓐ. Reversible process
ⓑ. Irreversible process
ⓒ. Isothermal adiabatic process
ⓓ. Non-spontaneous process
Correct Answer: Reversible process
Explanation: The melting of ice at its equilibrium temperature (0°C) is reversible because it occurs infinitesimally slowly and can be reversed by a tiny temperature change. The system and surroundings stay in equilibrium throughout, making the process quasi-static and reversible.
316. Which of the following conditions makes a process irreversible?
ⓐ. Occurs infinitely slowly with equilibrium at all stages
ⓑ. Produces no entropy
ⓒ. Occurs spontaneously with finite driving force and entropy generation
ⓓ. Has ΔS_total = 0
Correct Answer: Occurs spontaneously with finite driving force and entropy generation
Explanation: Irreversible processes happen spontaneously due to a finite difference in pressure, temperature, or concentration between system and surroundings. This non-equilibrium nature leads to entropy production \(ΔS_total > 0\), which distinguishes them from reversible ones.
317. In an isothermal reversible expansion of an ideal gas, the entropy change of the system is:
ⓐ. Zero
ⓑ. \(\Delta S = nR \ln \dfrac{V_2}{V_1}\)
ⓒ. \(\Delta S = nR \ln \dfrac{T_2}{T_1}\)
ⓓ. \(\Delta S = -nR \ln \dfrac{V_2}{V_1}\)
Correct Answer: \(\Delta S = nR \ln \dfrac{V_2}{V_1}\)
Explanation: For isothermal reversible expansion, heat absorbed equals work done:
\(q_{rev} = nRT \ln \dfrac{V_2}{V_1}\).
Dividing by temperature gives \(\Delta S = \dfrac{q_{rev}}{T} = nR \ln \dfrac{V_2}{V_1}\).
Since \(V_2 > V_1\), the logarithm is positive, meaning the system’s entropy increases.
318. For the same volume change, how does entropy change differ between reversible and irreversible expansion of a gas?
ⓐ. ΔS(system) is larger in irreversible expansion
ⓑ. ΔS(system) is the same, but ΔS(surroundings) differs
ⓒ. ΔS(system) is smaller in reversible expansion
ⓓ. ΔS(system) is negative in both
Correct Answer: ΔS(system) is the same, but ΔS(surroundings) differs
Explanation: Entropy is a state function, so ΔS_system depends only on initial and final states, not on the path. However, the surroundings lose less entropy in an irreversible expansion because less heat is exchanged reversibly. Hence, ΔS_universe increases in irreversible expansion.
319. In an irreversible process, the total entropy change (system + surroundings) is always:
ⓐ. Zero
ⓑ. Negative
ⓒ. Positive
ⓓ. Unpredictable
Correct Answer: Positive
Explanation: The total entropy of the universe always increases for irreversible processes. This principle represents the natural direction of spontaneous changes in the universe, ensuring the overall tendency toward disorder and equilibrium.
320. The second law of thermodynamics can be expressed in terms of entropy as:
ⓐ. ΔS_universe = 0 for irreversible processes
ⓑ. ΔS_universe ≤ 0 for all processes
ⓒ. ΔS_universe ≥ 0 for all spontaneous processes
ⓓ. ΔS_system = 0 always
Correct Answer: ΔS_universe ≥ 0 for all spontaneous processes
Explanation: The second law states that the entropy of the universe increases in any spontaneous process \(ΔS_universe > 0\) and remains constant for reversible ones \(ΔS_universe = 0\). This provides a universal criterion for determining spontaneity and equilibrium in thermodynamic systems.
321. What does the symbol ( G ) represent in thermodynamics?
ⓐ. Internal energy of the system
ⓑ. Gibbs free energy of the system
ⓒ. Heat content of the system
ⓓ. Entropy of the surroundings
Correct Answer: Gibbs free energy of the system
Explanation: Gibbs free energy (( G )) is a thermodynamic potential that combines enthalpy (( H )) and entropy (( S )) to determine whether a process is spontaneous at constant temperature and pressure. It is defined as \(G = H – TS\), where ( T ) is the absolute temperature.
322. What is the formula for Gibbs free energy?
ⓐ. \(G = H + TS\)
ⓑ. \(G = T – HS\)
ⓒ. \(G = U – PV\)
ⓓ. \(G = H – TS\)
Correct Answer: \(G = H – TS\)
Explanation: Gibbs free energy is defined by the relation \(G = H – TS\), where ( H ) is the enthalpy, ( T ) is the temperature in kelvin, and ( S ) is the entropy. It indicates the usable energy in a system capable of doing non-expansion (useful) work at constant temperature and pressure.
323. What does the term ( TS ) represent in the Gibbs free energy equation?
ⓐ. The total work done by the system
ⓑ. The energy associated with randomness or disorder
ⓒ. The heat content of the system
ⓓ. The kinetic energy of the molecules
Correct Answer: The energy associated with randomness or disorder
Explanation: The product ( TS ) represents the energy that is unavailable to do useful work because it is lost to randomness or disorder within the system. Subtracting ( TS ) from ( H ) in the equation \(G = H – TS\) gives the maximum useful work obtainable under constant temperature and pressure.
324. Which of the following best describes Gibbs free energy?
ⓐ. Total energy stored in the system
ⓑ. Energy available for doing useful work at constant temperature and pressure
ⓒ. Energy lost as heat during reaction
ⓓ. Energy required to break chemical bonds
Correct Answer: Energy available for doing useful work at constant temperature and pressure
Explanation: Gibbs free energy quantifies the portion of total energy (enthalpy) that can perform non-expansion work, such as electrical or chemical work, in processes at constant temperature and pressure. It is the most useful function for predicting reaction spontaneity and equilibrium.
325. If \(\Delta G = \Delta H – T\Delta S\), which condition indicates a spontaneous process?
ⓐ. \(\Delta G > 0\)
ⓑ. \(\Delta G = 0\)
ⓒ. \(\Delta G < 0\)
ⓓ. \(\Delta H > 0\)
Correct Answer: \(\Delta G < 0\)
Explanation: A process is spontaneous when Gibbs free energy decreases, i.e., \(\Delta G < 0\). This indicates that the system releases free energy and moves toward equilibrium. When \(\Delta G = 0\), the system is at equilibrium, and when \(\Delta G > 0\), the process is non-spontaneous.
326. Under which conditions does Gibbs free energy remain constant?
ⓐ. Constant volume and pressure
ⓑ. Constant temperature and pressure
ⓒ. Constant temperature and entropy
ⓓ. Constant internal energy
Correct Answer: Constant temperature and pressure
Explanation: The Gibbs function is particularly useful for processes occurring at constant temperature and pressure, such as most chemical reactions. Under these conditions, changes in Gibbs energy directly indicate spontaneity or equilibrium without requiring direct measurement of heat or work.
327. Which of the following equations relates Gibbs free energy to maximum work?
ⓐ. \(\Delta G = w_{max}\)
ⓑ. \(\Delta G = -T\Delta S\)
ⓒ. \(\Delta G = q_{rev}\)
ⓓ. \(\Delta G = -w_{max}\)
Correct Answer: \(\Delta G = -w_{max}\)
Explanation: The change in Gibbs free energy equals the negative of the maximum non-expansion work a system can perform under constant temperature and pressure. A negative \(\Delta G\) implies that the system releases energy available for doing useful work, such as in electrochemical cells.
328. If \(G = H – TS\), which factor decreases Gibbs free energy the most at high temperatures?
ⓐ. Increase in enthalpy
ⓑ. Decrease in temperature
ⓒ. Increase in entropy
ⓓ. Decrease in volume
Correct Answer: Increase in entropy
Explanation: In \(G = H – TS\), the ( TS ) term becomes significant at high temperatures. If entropy increases (ΔS > 0), the ( TΔS ) term grows larger and subtracts more from ( H ), decreasing ( G ). Therefore, endothermic reactions with large entropy increases often become spontaneous only at high temperatures.
329. For a reaction with \(\Delta H > 0\) and \(\Delta S > 0\), the process will be spontaneous when:
ⓐ. Temperature is low
ⓑ. ΔH = 0
ⓒ. Pressure is high
ⓓ. Temperature is high
Correct Answer: Temperature is high
Explanation: From \(\Delta G = \Delta H – T\Delta S\), when both ΔH and ΔS are positive, the process becomes spontaneous only at high temperatures, where \(T\Delta S > \Delta H\), making ΔG negative. Examples include melting and evaporation, which require heat but increase disorder significantly.
330. Why is Gibbs free energy an important thermodynamic function?
ⓐ. It predicts whether a process absorbs heat
ⓑ. It measures only the heat capacity of a system
ⓒ. It determines the spontaneity and maximum useful work obtainable from a process
ⓓ. It measures the randomness of a system
Correct Answer: It determines the spontaneity and maximum useful work obtainable from a process
Explanation: Gibbs free energy integrates both enthalpy and entropy changes to determine if a reaction can occur spontaneously under given conditions. A negative ΔG indicates spontaneous behavior, and its magnitude equals the maximum non-expansion work the system can perform, making it a central function in chemical thermodynamics.
331. What is the relationship between Gibbs free energy, enthalpy, and entropy?
ⓐ. \(\Delta G = \Delta H + T\Delta S\)
ⓑ. \(\Delta G = \Delta H – T\Delta S\)
ⓒ. \(\Delta G = \Delta H / T\Delta S\)
ⓓ. \(\Delta G = T\Delta S – \Delta H\)
Correct Answer: \(\Delta G = \Delta H – T\Delta S\)
Explanation: The Gibbs free energy equation combines enthalpy (ΔH) and entropy (ΔS) effects to determine spontaneity. It shows that a reaction becomes spontaneous (ΔG < 0) when the energy released as heat or increased disorder outweighs the energy required, making the process thermodynamically favorable.
332. If ΔH = –100 kJ and ΔS = +200 J/K, at what temperature does the reaction become spontaneous?
ⓐ. 100 K
ⓑ. 200 K
ⓒ. 300 K
ⓓ. 500 K
Correct Answer: 500 K
Explanation: A reaction is spontaneous when ΔG < 0. Converting ΔS to kJ/K:
ΔS = +0.2 kJ/K.
At equilibrium \(ΔG = 0\): \(T = \dfrac{\Delta H}{\Delta S} = \dfrac{100}{0.2} = 500 ,K\).
Thus, above 500 K, ΔG becomes negative, and the reaction is spontaneous at higher temperatures.
333. For a reaction where ΔH = +50 kJ and ΔS = +0.2 kJ/K, which statement is true?
ⓐ. The reaction is spontaneous at all temperatures
ⓑ. The reaction is spontaneous at low temperatures only
ⓒ. The reaction is spontaneous at high temperatures only
ⓓ. The reaction is never spontaneous
Correct Answer: The reaction is spontaneous at high temperatures only
Explanation: Here ΔH > 0 and ΔS > 0. At low T, ΔH dominates, giving ΔG > 0 (non-spontaneous). At high T, the term \(T\Delta S\) becomes large enough to make ΔG negative, resulting in spontaneous behavior at elevated temperatures.
334. For a process with ΔH = –200 kJ and ΔS = –0.3 kJ/K, when is the process spontaneous?
ⓐ. At all temperatures
ⓑ. Only at low temperatures
ⓒ. Only at high temperatures
ⓓ. Never spontaneous
Correct Answer: Only at low temperatures
Explanation: Here both ΔH and ΔS are negative. At low T, the –TΔS term is small, so ΔG remains negative (spontaneous). At high T, the –TΔS term becomes large and positive, turning ΔG positive and making the process non-spontaneous.
335. Which combination of ΔH and ΔS values always gives a non-spontaneous reaction?
ⓐ. ΔH < 0, ΔS > 0
ⓑ. ΔH < 0, ΔS < 0
ⓒ. ΔH > 0, ΔS > 0
ⓓ. ΔH > 0, ΔS < 0
Correct Answer: ΔH > 0, ΔS < 0
Explanation: When both enthalpy and entropy changes are unfavorable (ΔH positive and ΔS negative), ΔG = ΔH – TΔS is always positive at any temperature. Hence, the process cannot occur spontaneously under any condition.
336. A reaction is endothermic but spontaneous at room temperature. What does this suggest about ΔS?
ⓐ. ΔS is zero
ⓑ. ΔS is negative
ⓒ. ΔS is large and positive
ⓓ. ΔS is small and negative
Correct Answer: ΔS is large and positive
Explanation: Even though ΔH > 0 (endothermic), a process can still be spontaneous if the increase in entropy (ΔS > 0) is sufficiently large. When \(T\Delta S > \Delta H\), ΔG becomes negative, making the process thermodynamically favorable. Example: melting of ice at 298 K.
337. For an exothermic reaction with ΔS < 0, what determines whether it is spontaneous?
ⓐ. The value of ΔH only
ⓑ. The sign of ΔS only
ⓒ. The temperature at which it occurs
ⓓ. The total energy released
Correct Answer: The temperature at which it occurs
Explanation: If ΔH < 0 and ΔS < 0, spontaneity depends on temperature. At low T, ΔH dominates, and ΔG is negative (spontaneous). At high T, the –TΔS term becomes positive and may make ΔG positive, rendering the reaction non-spontaneous.
338. If a chemical reaction has ΔH = –150 kJ and ΔS = +0.1 kJ/K, what will be the sign of ΔG at 298 K?
ⓐ. Positive
ⓑ. Negative
ⓒ. Zero
ⓓ. Cannot be predicted
Correct Answer: Negative
Explanation: Using \(\Delta G = \Delta H – T\Delta S\):
\(\Delta G = (-150) – (298)(0.1) = -150 – 29.8 = -179.8 , kJ\).
Since ΔG < 0, the process is spontaneous at 298 K. The exothermic enthalpy and positive entropy both favor spontaneity.
339. When ΔG = 0, what does it signify?
ⓐ. The process is non-spontaneous
ⓑ. The process is spontaneous
ⓒ. The system is unstable
ⓓ. The system is at equilibrium
Correct Answer: The system is at equilibrium
Explanation: When ΔG = 0, the system has reached equilibrium. There is no net energy change, and the forward and reverse processes occur at equal rates. This balance marks the transition between spontaneous (ΔG < 0) and non-spontaneous (ΔG > 0) behavior.
340. What is the thermodynamic significance of a negative ΔG value?
ⓐ. The process absorbs energy from surroundings
ⓑ. The process is non-spontaneous
ⓒ. The process releases free energy and occurs spontaneously
ⓓ. The process is reversible but slow
Correct Answer: The process releases free energy and occurs spontaneously
Explanation: A negative ΔG indicates that the reaction proceeds spontaneously in the forward direction, releasing free energy available for work. It means that the total energy of the system decreases, driving the process naturally toward equilibrium.
341. What is the general thermodynamic criterion for spontaneity of a process at constant temperature and pressure?
ⓐ. \(\Delta H = 0\)
ⓑ. \(\Delta G < 0\)
ⓒ. \(\Delta S = 0\)
ⓓ. \(\Delta G > 0\)
Correct Answer: \(\Delta G < 0\)
Explanation: The Gibbs free energy change \(\Delta G\) determines spontaneity at constant temperature and pressure. A process is spontaneous when \(\Delta G < 0\), non-spontaneous when \(\Delta G > 0\), and at equilibrium when \(\Delta G = 0\). This is the most practical criterion for chemical reactions under laboratory and biological conditions.
342. What does a positive value of \(\Delta G\) indicate about a process?
ⓐ. It occurs spontaneously
ⓑ. It is at equilibrium
ⓒ. It is non-spontaneous under given conditions
ⓓ. It releases energy
Correct Answer: It is non-spontaneous under given conditions
Explanation: A positive \(\Delta G\) means the reaction requires energy input to proceed, making it thermodynamically unfavorable. However, if external conditions such as temperature or pressure change, the sign of \(\Delta G\) can reverse, allowing spontaneity.
343. Which of the following reactions is spontaneous at constant temperature and pressure?
ⓐ. \(\Delta G > 0\)
ⓑ. \(\Delta G = 0\)
ⓒ. \(\Delta G < 0\)
ⓓ. \(\Delta G = +T\Delta S\)
Correct Answer: \(\Delta G < 0\)
Explanation: The criterion \(\Delta G < 0\) corresponds to a decrease in Gibbs free energy, meaning the system releases free energy. The process will move naturally toward lower free energy until equilibrium is achieved, where \(\Delta G = 0\).
344. For a process with \(\Delta H < 0\) and \(\Delta S > 0\), what will be the sign of \(\Delta G\)?
ⓐ. Always positive
ⓑ. Zero
ⓒ. Depends on temperature
ⓓ. Always negative Zero
Correct Answer: Always negative
Explanation: In the equation \(\Delta G = \Delta H – T\Delta S\), both terms contribute to a negative value when \(\Delta H < 0\) and \(\Delta S > 0\). Hence, \(\Delta G\) remains negative at all temperatures, meaning the process is spontaneous under all conditions.
345. For an endothermic process to be spontaneous, what must be true?
ⓐ. \(\Delta H < 0\) and \(\Delta S < 0\)
ⓑ. \(\Delta H > 0\) and \(\Delta S > 0\)
ⓒ. \(\Delta H > 0\) and \(\Delta S < 0\)
ⓓ. \(\Delta H = 0\)
Correct Answer: \(\Delta H > 0\) and \(\Delta S > 0\)
Explanation: For endothermic processes \(\Delta H > 0\) to be spontaneous, entropy must increase significantly \(\Delta S > 0\) so that \(T\Delta S\) outweighs \(\Delta H\), making \(\Delta G\) negative. Examples include melting of ice and evaporation of water.
346. Which statement is correct for a process at equilibrium?
ⓐ. \(\Delta G < 0\)
ⓑ. \(\Delta G > 0\)
ⓒ. \(\Delta S = 0\)
ⓓ. \(\Delta G = 0\)
Correct Answer: \(\Delta G = 0\)
Explanation: At equilibrium, the forward and reverse reactions occur at the same rate, and no net change occurs in the system. The Gibbs free energy reaches a minimum value, and any small change increases it, thus preventing further spontaneous movement in either direction.
347. Which condition is necessary for a reaction to be spontaneous in the forward direction?
ⓐ. \(\Delta G > 0\)
ⓑ. \(\Delta G = 0\)
ⓒ. \(\Delta G < 0\)
ⓓ. \(\Delta H = 0\)
Correct Answer: \(\Delta G < 0\)
Explanation: The process will proceed spontaneously when the Gibbs free energy change is negative. This condition signifies that energy is released or available to perform work, driving the reaction naturally toward equilibrium.
348. Which of the following correctly describes a non-spontaneous process?
ⓐ. \(\Delta G < 0\) and proceeds naturally
ⓑ. \(\Delta G > 0\) and needs external energy to proceed
ⓒ. \(\Delta G = 0\) and is at equilibrium
ⓓ. \(\Delta S > 0\) and is always spontaneous
Correct Answer: \(\Delta G > 0\) and needs external energy to proceed
Explanation: Non-spontaneous processes are those that require external work or energy input. For example, the electrolysis of water occurs only when electrical energy is supplied, corresponding to a positive \(\Delta G\).
349. When will a reaction with \(\Delta H < 0\) and \(\Delta S < 0\) be spontaneous?
ⓐ. Only at high temperature
ⓑ. Only at low temperature
ⓒ. At all temperatures
ⓓ. Never spontaneous
Correct Answer: Only at low temperature
Explanation: For \(\Delta H < 0\) and \(\Delta S < 0\), the process can be spontaneous only when the temperature is low enough that \(T\Delta S\) is small. Then, \(\Delta G = \Delta H - T\Delta S\) remains negative. At higher temperatures, \(T\Delta S\) becomes large and positive, making \(\Delta G\) positive (non-spontaneous).
350. Which of the following statements is true about spontaneity based on Gibbs free energy?
ⓐ. All spontaneous processes have \(\Delta G > 0\)
ⓑ. Spontaneous processes may have positive or negative \(\Delta G\)
ⓒ. Spontaneous processes occur when \(\Delta G < 0\)
ⓓ. Spontaneous processes occur only when \(\Delta G = 0\)
Correct Answer: Spontaneous processes occur when \(\Delta G < 0\)
Explanation: The spontaneity of a process depends on the sign of the Gibbs free energy change. When \(\Delta G < 0\), the reaction proceeds spontaneously in the forward direction. When \(\Delta G > 0\), it is non-spontaneous, and \(\Delta G = 0\) corresponds to equilibrium.
351. What does the third law of thermodynamics state?
ⓐ. The entropy of a system always increases in a spontaneous process
ⓑ. The entropy of a perfect crystal at absolute zero is zero
ⓒ. The entropy of the universe always decreases
ⓓ. The entropy of gases is always constant at any temperature
Correct Answer: The entropy of a perfect crystal at absolute zero is zero
Explanation: The third law states that as temperature approaches absolute zero (0 K), the entropy of a perfect crystalline substance approaches zero. At this point, the atoms are in their most ordered state, with only one possible microstate, hence \(S = 0\).
352. What happens to the entropy of all pure crystalline solids at 0 K according to the third law?
ⓐ. It becomes infinite
ⓑ. It becomes negative
ⓒ. It remains constant but nonzero
ⓓ. It becomes zero
Correct Answer: It becomes zero
Explanation: At absolute zero, a perfect crystal has complete order and only one possible microstate, making its entropy zero. This provides a reference point for measuring absolute entropy of substances at higher temperatures.
353. Which of the following statements best explains the third law of thermodynamics?
ⓐ. It is impossible to reach absolute zero in a finite number of steps
ⓑ. Energy can neither be created nor destroyed
ⓒ. The entropy change of the universe is always positive
ⓓ. The entropy of an isolated system remains constant
Correct Answer: It is impossible to reach absolute zero in a finite number of steps
Explanation: The third law also implies that reaching 0 K is impossible because as temperature decreases, removing the last traces of heat requires infinite effort. This is consistent with the behavior of entropy approaching zero asymptotically.
354. According to the third law, what is the entropy of an imperfect crystal at 0 K?
ⓐ. Zero
ⓑ. Greater than zero
ⓒ. Negative
ⓓ. Infinite
Correct Answer: Greater than zero
Explanation: Imperfect crystals contain defects or disorder even at 0 K, giving them multiple microstates. Hence, entropy cannot reach zero. Only a perfect, defect-free crystal achieves zero entropy at absolute zero.
355. Which thermodynamic function becomes zero at absolute zero for a perfect crystal?
ⓐ. Internal energy
ⓑ. Gibbs free energy
ⓒ. Enthalpy
ⓓ. Entropy
Correct Answer: Entropy
Explanation: At 0 K, a perfect crystal has only one possible microstate (maximum order), so its entropy (S) = 0. This serves as the foundation for defining absolute entropy values of substances at higher temperatures through integration of heat capacities.
356. The third law of thermodynamics provides a method to determine which of the following?
ⓐ. Absolute values of entropy
ⓑ. Changes in enthalpy
ⓒ. Absolute values of enthalpy
ⓓ. Internal energy of gases
Correct Answer: Absolute values of entropy
Explanation: Since the entropy of a perfect crystal is defined as zero at 0 K, the absolute entropy at any temperature can be calculated by integrating heat capacity over temperature:
\(S = \int_0^T \dfrac{C_p}{T} dT\).
Thus, the third law gives an absolute scale for entropy.
357. Which of the following statements about entropy at 0 K is true?
ⓐ. Entropy is maximum
ⓑ. Entropy is minimum and equal to zero for perfect crystals
ⓒ. Entropy is undefined
ⓓ. Entropy equals heat capacity
Correct Answer: Entropy is minimum and equal to zero for perfect crystals
Explanation: A perfect crystal at 0 K has complete molecular order and a single microstate. Therefore, the entropy is minimum (zero). Any deviation from perfect order, such as defects or isotopic mixing, increases entropy above zero even near absolute zero.
358. The third law of thermodynamics is useful in calculating which of the following?
ⓐ. Entropy change between two finite temperatures
ⓑ. Latent heat of fusion
ⓒ. Enthalpy of vaporization
ⓓ. Heat of combustion
Correct Answer: Entropy change between two finite temperatures
Explanation: Using the third law as a baseline \(S = 0 at 0 K\), absolute entropy values can be determined for substances at higher temperatures. The difference between these gives the entropy change between two temperatures, vital in thermodynamic calculations.
359. Why is it impossible to reach absolute zero temperature?
ⓐ. Because matter loses all its mass
ⓑ. Because removing all heat requires infinite steps or work
ⓒ. Because entropy increases continuously
ⓓ. Because all energy disappears
Correct Answer: Because removing all heat requires infinite steps or work
Explanation: According to the third law, the closer a system gets to absolute zero, the harder it becomes to extract additional heat. The cooling rate slows down infinitely, making 0 K unattainable in finite time or by any finite number of steps.
360. Which of the following correctly summarizes the third law of thermodynamics?
ⓐ. The entropy of a perfect crystal approaches zero as temperature approaches absolute zero
ⓑ. The energy of an ideal gas is zero at 0 K
ⓒ. The entropy of the universe is always increasing
ⓓ. The enthalpy of a perfect gas is constant at 0 K
Correct Answer: The entropy of a perfect crystal approaches zero as temperature approaches absolute zero
Explanation: The third law establishes the zero point of entropy. As temperature approaches 0 K, molecular motion nearly ceases, and the system becomes perfectly ordered with S → 0. This concept enables calculation of absolute entropies and helps define temperature on an absolute scale.
361. What is the absolute entropy of a perfect crystalline solid at 0 K?
ⓐ. Infinite
ⓑ. Negative
ⓒ. Constant but nonzero
ⓓ. Zero
Correct Answer: Zero
Explanation: According to the third law of thermodynamics, the absolute entropy of a perfect crystalline solid at 0 K is zero because there is complete order and only one possible microstate. With no randomness or disorder, the entropy value becomes exactly zero, serving as a reference for calculating absolute entropies at higher temperatures.
362. Why does the absolute entropy of a perfect crystal become zero at 0 K?
ⓐ. The internal energy becomes infinite
ⓑ. The atoms stop vibrating completely and occupy fixed positions
ⓒ. The heat capacity becomes infinite
ⓓ. The system loses all its mass
Correct Answer: The atoms stop vibrating completely and occupy fixed positions
Explanation: At absolute zero, atomic motion ceases, leaving only one microstate possible, corresponding to perfect order. Since entropy measures the number of microstates \(S = k \ln W\), and \(W = 1\), \(S = 0\). Thus, there is no disorder in a perfect crystal at 0 K.
363. Which of the following correctly expresses the entropy at 0 K for a perfect crystal?
ⓐ. \(S = k \ln 0\)
ⓑ. \(S = 0\)
ⓒ. \(S = \infty\)
ⓓ. \(S = k \ln 1\)
Correct Answer: \(S = k \ln 1\)
Explanation: Entropy is defined by Boltzmann’s relation \(S = k \ln W\), where ( W ) is the number of microstates. For a perfect crystal at 0 K, \(W = 1\), so \(S = k \ln 1 = 0\). This mathematical expression supports the third law of thermodynamics, confirming zero entropy at absolute zero.
364. Which of the following statements about absolute entropy is true?
ⓐ. It is always negative at 0 K
ⓑ. It depends on the path taken
ⓒ. It is defined as zero for a perfect crystal at absolute zero
ⓓ. It increases infinitely as temperature decreases
Correct Answer: It is defined as zero for a perfect crystal at absolute zero
Explanation: The third law defines absolute entropy as zero at 0 K for a perfectly ordered crystal. This forms the basis for calculating absolute entropy at higher temperatures using \(S = \int_0^T \frac{C_p}{T} dT\). It is a state function and independent of the process path.
365. Why does an imperfect crystal have nonzero entropy at 0 K?
ⓐ. It has more than one possible microstate due to structural defects
ⓑ. It releases latent heat
ⓒ. It contains no atoms in motion
ⓓ. It expands indefinitely
Correct Answer: It has more than one possible microstate due to structural defects
Explanation: Defects such as dislocations or impurities cause partial disorder even at absolute zero. Because multiple arrangements (microstates) are possible, entropy remains greater than zero. Thus, only a perfect crystal achieves S = 0 at 0 K, while imperfect crystals retain residual entropy.
366. What is the term used for the remaining entropy in a substance due to disorder at 0 K?
ⓐ. Total entropy
ⓑ. Residual entropy
ⓒ. Lattice energy
ⓓ. Absolute enthalpy
Correct Answer: Residual entropy
Explanation: Residual entropy is the entropy retained at 0 K because of molecular disorder, even in the absence of thermal motion. For example, CO crystals have residual entropy due to random orientations of CO molecules, which prevents complete ordering even at absolute zero.
367. How can the absolute entropy of a substance at any temperature be calculated using the third law?
ⓐ. \(S = \int_T^{0} \frac{C_p}{T} dT\)
ⓑ. \(S = C_p / T\)
ⓒ. \(S = C_p \times T\)
ⓓ. \(S = \int_0^T \frac{C_p}{T} dT\)
Correct Answer: \(S = \int_0^T \frac{C_p}{T} dT\)
Explanation: The third law provides a reference point \(S = 0 at 0 K\) to compute absolute entropy at higher temperatures by integrating the ratio of heat capacity \(C_p\) to temperature from 0 K to T. This gives the total entropy gained as the system warms.
368. The absolute entropy of gases is higher than that of solids because:
ⓐ. Gases have lower molecular weight
ⓑ. Gases have more possible microstates and higher disorder
ⓒ. Gases have fixed positions of particles
ⓓ. Gases have no internal energy
Correct Answer: Gases have more possible microstates and higher disorder
Explanation: Entropy depends on molecular randomness. In gases, particles are free to move and occupy a vast number of configurations. Thus, gases have the highest absolute entropy, followed by liquids, then solids. This increasing disorder explains the difference in entropy among states.
369. Which of the following is true for entropy at absolute zero according to the third law?
ⓐ. The entropy of a substance is infinite
ⓑ. The entropy of gases becomes constant
ⓒ. The entropy of a solid solution is negative
ⓓ. The entropy of a perfect crystal is zero
Correct Answer: The entropy of a perfect crystal is zero
Explanation: The third law of thermodynamics clearly states that the entropy of a perfect crystal approaches zero as the temperature approaches absolute zero. This establishes the zero-point reference for entropy measurements across all substances.
370. Why can the third law be used to determine absolute entropies experimentally?
ⓐ. Because heat capacity can be integrated from 0 K to T
ⓑ. Because enthalpy is constant at all temperatures
ⓒ. Because entropy is independent of temperature
ⓓ. Because energy becomes infinite at 0 K
Correct Answer: Because heat capacity can be integrated from 0 K to T
Explanation: Using the third law, the absolute entropy at temperature T can be obtained by integrating the heat capacity ratio \(\frac{C_p}{T}\) from 0 K upward. This experimental method relies on measurable heat capacities and provides accurate absolute entropy values for real substances.
371. Which of the following is a major application of the third law of thermodynamics?
ⓐ. Determining absolute enthalpy
ⓑ. Calculating absolute entropy of substances
ⓒ. Measuring internal energy directly
ⓓ. Finding critical temperature of gases
Correct Answer: Calculating absolute entropy of substances
Explanation: The third law provides a zero-entropy reference at 0 K, allowing calculation of absolute entropies at higher temperatures using \(S = \int_0^T \frac{C_p}{T} dT\). These values are essential for predicting reaction feasibility, equilibrium constants, and phase stability in thermodynamics.
372. How does the third law help in calculating entropy changes for chemical reactions?
ⓐ. By defining zero enthalpy at 0 K
ⓑ. By eliminating the need for heat capacity data
ⓒ. By giving a direct value for ΔH
ⓓ. By allowing the use of absolute entropies of reactants and products
Correct Answer: By allowing the use of absolute entropies of reactants and products
Explanation: The third law provides a foundation for computing reaction entropy changes:
\(\Delta S_{reaction} = \sum S_{products} – \sum S_{reactants}\).
Absolute entropies for each species are determined using the third-law reference \(S = 0 at 0 K\), enabling accurate calculations for reactions under standard conditions.
373. Which of the following processes can be better understood using the third law of thermodynamics?
ⓐ. Radioactive decay
ⓑ. Phase transitions such as melting and vaporization
ⓒ. Elastic deformation of solids
ⓓ. Electrical conductivity in metals
Correct Answer: Phase transitions such as melting and vaporization
Explanation: The third law helps analyze phase transitions by providing absolute entropy values for each phase. The entropy change during transitions like melting or vaporization can be computed using \(\Delta S = \frac{\Delta H_{transition}}{T_{transition}}\), aiding in predicting equilibrium and spontaneity.
374. The third law of thermodynamics helps explain which experimental observation?
ⓐ. It is impossible to reach absolute zero temperature
ⓑ. Gases expand on heating
ⓒ. Heat flows from cold to hot
ⓓ. Pressure of gases increases with temperature
Correct Answer: It is impossible to reach absolute zero temperature
Explanation: As temperature approaches 0 K, removing the last traces of heat becomes increasingly difficult. The entropy change approaches zero, and cooling efficiency declines, requiring infinite steps to reach 0 K. Hence, absolute zero is practically unattainable.
375. Why is the third law useful in cryogenic studies (very low-temperature research)?
ⓐ. It predicts energy levels of gases
ⓑ. It explains chemical kinetics
ⓒ. It helps in determining the behavior of substances near absolute zero
ⓓ. It defines vapor pressure constants
Correct Answer: It helps in determining the behavior of substances near absolute zero
Explanation: The third law allows scientists to understand how entropy, heat capacity, and phase changes behave near 0 K. This is crucial for studying superconductors, liquefied gases, and quantum phenomena occurring in extremely low-temperature environments.
376. How can the third law be applied to determine the feasibility of a reaction?
ⓐ. By measuring pressure changes
ⓑ. By analyzing volume changes
ⓒ. By calculating internal energy directly
ⓓ. By using absolute entropies to find ΔG through ΔH – TΔS
Correct Answer: By using absolute entropies to find ΔG through ΔH – TΔS
Explanation: Once absolute entropies are known from the third law, the Gibbs free energy change can be determined using \(\Delta G = \Delta H – T\Delta S\). The sign of ΔG reveals whether the reaction is spontaneous (ΔG < 0) or non-spontaneous (ΔG > 0).
377. In which way does the third law assist in determining chemical equilibrium constants?
ⓐ. By setting ΔG = 0 at all temperatures
ⓑ. By enabling accurate entropy calculations to find ΔG°
ⓒ. By defining the enthalpy of reaction as zero
ⓓ. By ignoring temperature effects on equilibrium
Correct Answer: By enabling accurate entropy calculations to find ΔG°
Explanation: Using \(\Delta G° = \Delta H° – T\Delta S°\) and entropy values obtained via the third law, equilibrium constants (( K )) can be computed from \(\Delta G° = -RT \ln K\). Thus, the third law indirectly helps predict the extent of reactions under standard conditions.
378. Why is residual entropy important in applying the third law?
ⓐ. It ensures entropy is always zero
ⓑ. It corrects for imperfections in real crystals when calculating entropy
ⓒ. It eliminates the need for experimental data
ⓓ. It makes all solids behave ideally
Correct Answer: It corrects for imperfections in real crystals when calculating entropy
Explanation: Residual entropy, caused by atomic disorder at 0 K, leads to slight deviations from the ideal zero-entropy condition. Considering this correction allows for more accurate thermodynamic calculations, especially in systems with orientational or structural defects.
379. Which property cannot be determined without the third law of thermodynamics?
ⓐ. Relative enthalpy
ⓑ. Heat of reaction
ⓒ. Change in pressure
ⓓ. Absolute entropy
Correct Answer: Absolute entropy
Explanation: While changes in enthalpy or pressure can be determined without the third law, absolute entropy values require the zero-entropy reference provided by it. This makes the third law indispensable for assigning precise entropy values to substances at any temperature.
380. Why is the third law essential in defining the entropy scale?
ⓐ. It fixes the reference point of entropy at 0 K
ⓑ. It provides the value of R (gas constant)
ⓒ. It defines the scale for enthalpy changes
ⓓ. It eliminates the need for temperature measurements
Correct Answer: It fixes the reference point of entropy at 0 K
Explanation: The third law establishes that the entropy of a perfect crystal is zero at absolute zero. This gives an absolute reference point for entropy, allowing thermodynamic quantities to be measured on a universal scale and ensuring consistent calculations across all substances and conditions.
381. One mole of an ideal gas \(n = 1.5 mol\) expands isothermally and reversibly at 300 K from 5.0 L to 15.0 L. What is the work done by the gas?
ⓐ. +4.11 kJ
ⓑ. −2.74 kJ
ⓒ. −4.11 kJ
ⓓ. −8.31 kJ
Correct Answer: −4.11 kJ
Explanation: For isothermal reversible expansion, \(w = -nRT\ln\left(\dfrac{V_2}{V_1}\right). Substituting \(n=1.5\), \(R=8.314,\mathrm{J,mol^{-1}K^{-1}}\), \(T=300,\mathrm{K}\), \(V_2/V_1=3\) gives \(w=-1.5\times 8.314\times 300\times \ln 3 \approx -4110,\mathrm{J}=-4.11,\mathrm{kJ}\). Negative sign indicates work done by the system on surroundings. Options with positive work correspond to compression or sign mistakes.
382. For a reaction at 298 K with \(\Delta H = -95,\mathrm{kJ,mol^{-1}}\) and \(\Delta S = -120,\mathrm{J,mol^{-1}K^{-1}}\), what is \(\Delta G\)?
ⓐ. −131 kJ mol⁻¹
ⓑ. −59.2 J mol⁻¹
ⓒ. +59.2 kJ mol⁻¹
ⓓ. −59.2 kJ mol⁻¹
Correct Answer: −59.2 kJ mol⁻¹
Explanation: Use \(\Delta G=\Delta H-T\Delta S\), converting entropy to kJ: \(\Delta S=-0.120,\mathrm{kJ,mol^{-1}K^{-1}}\). Then \(\Delta G=-95-298(-0.120)= -95+35.76=-59.24,\mathrm{kJ,mol^{-1}}\). Negative \(\Delta G\) indicates spontaneity despite a decrease in entropy, because the exothermic enthalpy drives the process.
383. At 298 K a reaction has \(\Delta G^\circ = +12.0,\mathrm{kJ,mol^{-1}}\). What is the equilibrium constant ( K )?
ⓐ. \(7.9\times 10^{-3}\)
ⓑ. \(3.2\times 10^{-2}\)
ⓒ. \(1.0\times 10^{-1}\)
ⓓ. \(2.5\times 10^{-3}\)
Correct Answer: \(7.9\times 10^{-3}\)
Explanation: \(\Delta G^\circ = -RT\ln K \Rightarrow K=\exp!\left(-\dfrac{\Delta G^\circ}{RT}\right). With \(R=8.314,\mathrm{J,mol^{-1}K^{-1}}\) and \(T=298,\mathrm{K}\), \(K=\exp!\left(-12000/(8.314\times 298)\right)\approx 7.9\times 10^{-3}\). A small ( K ) is consistent with positive \(\Delta G^\circ\).
384. An ideal gas has \(C_p = 29.1,\mathrm{J,mol^{-1}K^{-1}}\). What are \(C_v\) and \(\gamma\) for the gas?
ⓐ. \(C_v=20.8,\mathrm{J,mol^{-1}K^{-1}},\ \gamma=1.40\)
ⓑ. \(C_v=12.5,\mathrm{J,mol^{-1}K^{-1}},\ \gamma=2.33\)
ⓒ. \(C_v=25.8,\mathrm{J,mol^{-1}K^{-1}},\ \gamma=1.13\)
ⓓ. \(C_v=37.4,\mathrm{J,mol^{-1}K^{-1}},\ \gamma=0.78\)
Correct Answer: \(C_v=20.8,\mathrm{J,mol^{-1}K^{-1}},\ \gamma=1.40\)
Explanation: For an ideal gas, \(C_p-C_v=R=8.314\). Thus \(C_v=29.1-8.314=20.786\approx 20.8\). The heat capacity ratio \(\gamma=C_p/C_v=29.1/20.786\approx 1.40\), typical of diatomic gases at room temperature where rotational modes are active.
385. A diatomic ideal gas undergoes a reversible adiabatic expansion from \(2.0,\mathrm{L}\) at \(10,\mathrm{atm}\) to \(5.0,\mathrm{L}\). Take \(\gamma=1.40\). What is the final pressure?
ⓐ. 1.6 atm
ⓑ. 2.2 atm
ⓒ. 2.77 atm
ⓓ. 4.0 atm
Correct Answer: 2.77 atm
Explanation: For adiabatic reversible change, \(PV^\gamma=\text{constant}\). Hence \(P_2=P_1\left(\dfrac{V_1}{V_2}\right)^\gamma = 10,(2/5)^{1.4}\approx 2.77,\mathrm{atm}\). Choices far from 2.8 atm reflect misuse of \(\gamma\) or isothermal relation.
386. Using the data of question 385, if the initial temperature is 300 K, what is the final temperature?
ⓐ. 262 K
ⓑ. 241 K
ⓒ. 236 K
ⓓ. 208 K
Correct Answer: 208 K
Explanation: For a reversible adiabatic, \(TV^{\gamma-1}=\text{constant}\). Thus \(T_2=T_1\left(\dfrac{V_1}{V_2}\right)^{\gamma-1}=300\times (2/5)^{0.4}\approx 208,\mathrm{K}\). Temperature drops because expansion does work with no heat input.
387. A 50 g copper block at 100°C is dropped into 100 g of water at 25°C in an insulated beaker. Given \(c_\text{Cu}=0.385,\mathrm{J,g^{-1}K^{-1}}\), \(c_\text{water}=4.18,\mathrm{J,g^{-1}K^{-1}}\). What is the final temperature?
ⓐ. 27.1°C
ⓑ. 28.3°C
ⓒ. 30.0°C
ⓓ. 32.5°C
Correct Answer: 28.3°C
Explanation: Energy balance: \(m_wc_w(T_f-25)+m_\text{Cu}c_\text{Cu}(T_f-100)=0\). Solve \(100\cdot4.18+50\cdot0.385)T_f=100\cdot4.18\cdot25+50\cdot0.385\cdot100\). This yields \(T_f\approx 28.3^\circ\mathrm{C}\). Water’s large heat capacity keeps the final temperature close to 25°C.
388. Using formation enthalpies \(\Delta H_f^\circ(\mathrm{CaCO_3,s})=-1206.9,\mathrm{kJ,mol^{-1}}\), \(\Delta H_f^\circ(\mathrm{CaO,s})=-635.1\), \(\Delta H_f^\circ(\mathrm{CO_2,g})=-393.5\), compute \(\Delta H^\circ\) for \(\mathrm{CaCO_3(s)\rightarrow CaO(s)+CO_2(g)}\).
ⓐ. +178.3 kJ mol⁻¹
ⓑ. −178.3 kJ mol⁻¹
ⓒ. +241.8 kJ mol⁻¹
ⓓ. −241.8 kJ mol⁻¹
Correct Answer: +178.3 kJ mol⁻¹
Explanation: \(\Delta H^\circ = [\Delta H_f^\circ(\mathrm{CaO})+\Delta H_f^\circ(\mathrm{CO_2})]-\Delta H_f^\circ(\mathrm{CaCO_3})=(-635.1-393.5)-(-1206.9)=+178.3,\mathrm{kJ,mol^{-1}}\). Positive value shows thermal input is required to decompose limestone (endothermic calcination).
389. One mole of an ideal gas undergoes isothermal, reversible compression at 300 K from 10.0 L to 2.0 L. Compute the work on the gas.
ⓐ. +4.01 kJ
ⓑ. −4.01 kJ
ⓒ. +2.48 kJ
ⓓ. −2.48 kJ
Correct Answer: +4.01 kJ
Explanation: \(w=-nRT\ln(V_2/V_1)=-1\times 8.314\times 300\times \ln(0.2)=+4.01,\mathrm{kJ}\). Positive sign indicates work done on the system during compression. The magnitude mirrors the expansion work between the same states.
390. At 298 K, the equilibrium constant for a reaction is \(K=4.0\times 10^5\). What is \(\Delta G^\circ\)?
ⓐ. −18.3 kJ mol⁻¹
ⓑ. −26.5 kJ mol⁻¹
ⓒ. −32.0 kJ mol⁻¹
ⓓ. −41.1 kJ mol⁻¹
Correct Answer: −32.0 kJ mol⁻¹
Explanation: \(\Delta G^\circ=-RT\ln K\). With \(R=8.314,\mathrm{J,mol^{-1}K^{-1}}\), \(T=298,\mathrm{K}\), \(\ln K=\ln(4\times 10^5)\approx 12.9\). Thus \(\Delta G^\circ\approx -8.314\times 298\times 12.9\approx -3.20\times 10^4,\mathrm{J,mol^{-1}}=-32.0,\mathrm{kJ,mol^{-1}}\). Large ( K ) matches negative \(\Delta G^\circ\).
391. For an ideal gas with \(C_p=29.1,\mathrm{J,mol^{-1}K^{-1}}\) at 300 K, what is \(\gamma\) and \(C_v\)?
ⓐ. \(\gamma=1.67,\ C_v=17.4,\mathrm{J,mol^{-1}K^{-1}}\)
ⓑ. \(\gamma=1.40,\ C_v=20.8,\mathrm{J,mol^{-1}K^{-1}}\)
ⓒ. \(\gamma=1.33,\ C_v=21.9,\mathrm{J,mol^{-1}K^{-1}}\)
ⓓ. \(\gamma=1.25,\ C_v=23.3,\mathrm{J,mol^{-1}K^{-1}}\)
Correct Answer: \(\gamma=1.40,\ C_v=20.8,\mathrm{J,mol^{-1}K^{-1}}\)
Explanation: \(C_v=C_p-R=29.1-8.314=20.786\). Hence \(\gamma=C_p/C_v=29.1/20.786\approx 1.40\). Values align with diatomic gas behavior at room temperature where vibrational modes are mostly frozen out.
392. Two moles of an ideal gas are heated at constant pressure from 300 K to 500 K. If \(C_p=29.1,\mathrm{J,mol^{-1}K^{-1}}\), what is the entropy change of the gas?
ⓐ. 14.9 J K⁻¹
ⓑ. 20.1 J K⁻¹
ⓒ. 29.7 J K⁻¹
ⓓ. 41.5 J K⁻¹
Correct Answer: 29.7 J K⁻¹
Explanation: For \(P=\text{const}\), \(\Delta S = n\int_{T_1}^{T_2}\dfrac{C_p}{T}\\,dT = nC_p\ln\left(\dfrac{T_2}{T_1}\right). Thus \(\Delta S=2\times 29.1\times \ln(500/300)\approx 29.7,\mathrm{J,K^{-1}}\). Entropy rises because temperature and molecular randomness increase.
393. One mole of an ideal gas expands isothermally and reversibly at 298 K from 2.0 L to 8.0 L. What is \(\Delta S\) of the gas?
ⓐ. 5.76 J K⁻¹
ⓑ. 11.53 J K⁻¹
ⓒ. 18.53 J K⁻¹
ⓓ. 19.62 J K⁻¹
Correct Answer: 11.53 J K⁻¹
Explanation: \(\Delta S = nR\ln\left(\dfrac{V_2}{V_1}\right) = 1\times 8.314\times \ln(4) \approx 8.314\times 1.386=11.53,\mathrm{J,K^{-1}}\). Entropy increases due to greater volume and accessible microstates.
394. Three moles of a gas with temperature-independent \(C_p=75.3,\mathrm{J,mol^{-1}K^{-1}}\) are heated from 300 K to 350 K at constant pressure. What is \(\Delta H\)?
ⓐ. 7.53 kJ
ⓑ. 9.42 kJ
ⓒ. 10.1 kJ
ⓓ. 11.3 kJ
Correct Answer: 11.3 kJ
Explanation: At constant pressure, \(\Delta H = n C_p \Delta T = 3\times 75.3\times 50=11295,\mathrm{J}=11.3,\mathrm{kJ}\). This linear relation is valid when \(C_p\) is approximately constant over the temperature range.
395. For a reaction with \(\Delta H = +80.0,\mathrm{kJ,mol^{-1}}\) and \(\Delta S = +220,\mathrm{J,mol^{-1}K^{-1}}\), above what temperature will it be spontaneous?
ⓐ. 273 K
ⓑ. 298 K
ⓒ. 318 K
ⓓ. 364 K
Correct Answer: 364 K
Explanation: Spontaneity requires \(\Delta G=\Delta H – T\Delta S < 0 \Rightarrow T>\Delta H/\Delta S\). Convert units: \(\Delta S=0.220,\mathrm{kJ,mol^{-1}K^{-1}}\). Then \(T_\text{min}=80.0/0.220\approx 363.6,\mathrm{K}\). Hence temperatures above about 364 K make \(\Delta G\) negative.
You’re on Class 11 Chemistry MCQs – Chapter 6: Thermodynamics (Part 4), the last practice block for this chapter. Short, focused explanations keep you ready for Boards, JEE, NEET.
Focus of Part 4 (95 MCQs): Second Law & Entropy (ΔS); order–disorder ideas; ΔS in phase changes/mixing; Gibbs free energy (ΔG = ΔH − TΔS), spontaneity/feasibility at different temperatures, and common exam traps. Ideal for final concept polish + speed.
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