301. Entropy \(S\) is best understood as a thermodynamic state function related to
ⓐ. energy dispersal and randomness
ⓑ. only the colour intensity of a substance
ⓒ. the total mass of the laboratory table
ⓓ. pressure-volume work in every process only
Correct Answer: energy dispersal and randomness
Explanation: Entropy \(S\) is a thermodynamic state function connected with energy dispersal and molecular disorder. A gas usually has higher entropy than a liquid of the same substance, and a liquid usually has higher entropy than a solid. Entropy is not judged from colour alone. It is also not the same as pressure-volume work, although both are thermodynamic ideas. Since \(S\) is a state function, \(\Delta S\) depends on the initial and final states. The concept becomes especially useful for predicting the direction of spontaneous change.
302. For the same substance at the same temperature range, the arrangement with the greatest entropy is generally
ⓐ. solid
ⓑ. liquid
ⓒ. gas
ⓓ. perfectly ordered crystal at \(0\,\text{K}\)
Correct Answer: gas
Explanation: Gases generally have much greater freedom of motion than liquids and solids. Their particles are farther apart and can occupy a much larger volume, so the number of possible arrangements is greater. Liquids have more disorder than solids but less than gases. A perfectly ordered crystal at \(0\,\text{K}\) represents a very low-entropy limiting case. The order is usually \(S_{\text{gas}} \gt S_{\text{liquid}} \gt S_{\text{solid}}\) for the same substance under comparable conditions. This comparison is based on particle freedom, not on chemical formula alone.
303. Melting of ice at its melting point is accompanied by an increase in entropy because
ⓐ. water molecules stop moving completely in the liquid
ⓑ. liquid water has more accessible arrangements
ⓒ. the chemical formula changes from \(\mathrm{H_2O}\) to \(\mathrm{H_2O_2}\)
ⓓ. entropy must decrease in every phase change
Correct Answer: liquid water has more accessible arrangements
Explanation: During melting, solid ice changes into liquid water while the chemical identity remains \(\mathrm{H_2O}\). Liquid water allows more molecular movement and more possible arrangements than the ordered solid lattice. Therefore, the entropy of the system increases. The process does not convert water into \(\mathrm{H_2O_2}\). Entropy change depends on the direction and nature of the phase change; it does not always decrease. Melting is a common example where increased molecular freedom gives \(\Delta S \gt 0\).
304. The entropy change for a reversible heat transfer at temperature \(T\) is given by
ⓐ. \(\Delta S=q_{\text{rev}}T\)
ⓑ. \(\Delta S=\frac{q_{\text{rev}}}{T}\)
ⓒ. \(\Delta S=\frac{T}{q_{\text{rev}}}\)
ⓓ. \(\Delta S=p\Delta V\)
Correct Answer: \(\Delta S=\frac{q_{\text{rev}}}{T}\)
Explanation: Entropy change for a reversible heat transfer is defined by \(\Delta S=\frac{q_{\text{rev}}}{T}\). The heat must be reversible heat, and the temperature must be expressed in \(\text{K}\). This relation shows that the same heat transfer produces a larger entropy change at lower temperature. The expression \(p\Delta V\) is connected with pressure-volume work, not entropy. Entropy has units such as \(\text{J K}^{-1}\) or \(\text{J mol}^{-1}\text{K}^{-1}\). The temperature in the denominator is essential for the unit and physical meaning.
305. A reversible process absorbs \(600\,\text{J}\) of heat at \(300\,\text{K}\). The entropy change of the system is
ⓐ. \(-2.0\,\text{J K}^{-1}\)
ⓑ. \(+180000\,\text{J K}^{-1}\)
ⓒ. \(0\,\text{J K}^{-1}\)
ⓓ. \(+2.0\,\text{J K}^{-1}\)
Correct Answer: \(+2.0\,\text{J K}^{-1}\)
Explanation: \( \textbf{Given heat:} \) The system absorbs heat, so \(q_{\text{rev}}=+600\,\text{J}\).
\( \textbf{Temperature:} \) \(T=300\,\text{K}\).
\( \textbf{Entropy relation:} \)
\[
\Delta S=\frac{q_{\text{rev}}}{T}
\]
\( \textbf{Substitution:} \)
\[
\Delta S=\frac{+600\,\text{J}}{300\,\text{K}}
\]
\( \textbf{Calculation:} \)
\[
\Delta S=+2.0\,\text{J K}^{-1}
\]
\( \textbf{Sign meaning:} \) Positive entropy change matches heat absorption in this reversible step.
\( \textbf{Final answer:} \) \(\Delta S=+2.0\,\text{J K}^{-1}\). Dividing by absolute temperature gives entropy units, not energy units.
306. A system loses \(900\,\text{J}\) of heat reversibly at \(300\,\text{K}\). The entropy change of the system is
ⓐ. \(+3.0\,\text{J K}^{-1}\)
ⓑ. \(-600\,\text{J K}^{-1}\)
ⓒ. \(0\,\text{J K}^{-1}\)
ⓓ. \(-3.0\,\text{J K}^{-1}\)
Correct Answer: \(-3.0\,\text{J K}^{-1}\)
Explanation: \( \textbf{Heat direction:} \) The system loses heat, so \(q_{\text{rev}}=-900\,\text{J}\).
\( \textbf{Temperature:} \) \(T=300\,\text{K}\).
\( \textbf{Use the entropy formula:} \)
\[
\Delta S=\frac{q_{\text{rev}}}{T}
\]
\( \textbf{Substitute values:} \)
\[
\Delta S=\frac{-900\,\text{J}}{300\,\text{K}}
\]
\( \textbf{Simplify:} \)
\[
\Delta S=-3.0\,\text{J K}^{-1}
\]
\( \textbf{Interpretation:} \) Heat leaving the system in this reversible transfer lowers the system entropy.
\( \textbf{Final answer:} \) \(\Delta S=-3.0\,\text{J K}^{-1}\). The sign comes from the heat direction for the chosen system.
307. The unit most suitable for molar entropy is
ⓐ. \(\text{J mol K}\)
ⓑ. \(\text{mol J}^{-1}\text{K}^{-1}\)
ⓒ. \(\text{J mol}^{-1}\text{K}^{-1}\)
ⓓ. \(\text{K mol}^{-1}\)
Correct Answer: \(\text{J mol}^{-1}\text{K}^{-1}\)
Explanation: Entropy has the unit of energy divided by temperature, such as \(\text{J K}^{-1}\). When entropy is reported per mole, the unit becomes \(\text{J mol}^{-1}\text{K}^{-1}\). This is common for standard molar entropy values. The unit \(\text{J mol K}\) incorrectly places \(\text{mol}\) and \(\text{K}\) in the numerator. The unit also distinguishes entropy from enthalpy, which is often reported in \(\text{kJ mol}^{-1}\). Entropy calculations often require careful conversion because \(\Delta H\) may be in \(\text{kJ mol}^{-1}\) while entropy is in \(\text{J mol}^{-1}\text{K}^{-1}\).
308. A reaction produces more moles of gas than it consumes. Other factors being similar, the entropy change of the system is likely to be
ⓐ. negative
ⓑ. exactly zero in every case
ⓒ. positive
ⓓ. impossible because gases have no entropy
Correct Answer: positive
Explanation: Gases have large entropy because their particles have high freedom of motion and many possible arrangements. If a reaction produces more gaseous moles than it consumes, the disorder and energy dispersal of the system often increase. Therefore, \(\Delta S_{\text{system}}\) is likely to be positive. This is a qualitative prediction, so the exact value still depends on the substances and conditions. Gases certainly have entropy; in fact, gas formation often increases entropy strongly. The mole-count idea is useful when comparing reactions with clear changes in gaseous species.
309. For the reaction \(\mathrm{CaCO_3(s)\rightarrow CaO(s)+CO_2(g)}\), the sign of \(\Delta S_{\text{system}}\) is expected to be
ⓐ. negative because all products are solids
ⓑ. zero because the reaction has one reactant formula
ⓒ. negative because \(\mathrm{CO_2}\) contains oxygen
ⓓ. positive because a gaseous product is formed
Correct Answer: positive because a gaseous product is formed
Explanation: The reactant \(\mathrm{CaCO_3(s)}\) is a solid. The products include \(\mathrm{CO_2(g)}\), which has much greater entropy than a solid species. Formation of a gas from a solid reactant usually increases the number of accessible arrangements of the system. Therefore, \(\Delta S_{\text{system}}\) is expected to be positive. The number of written formulas alone is not the deciding factor. The physical state of the substances, especially gas formation, gives the main entropy clue here.
310. Study the table and choose the change most likely to have \(\Delta S_{\text{system}} \gt 0\).
| Row | Change |
|---|
| P | \(\mathrm{H_2O(g)\rightarrow H_2O(l)}\) |
| Q | \(\mathrm{NaCl(aq)\rightarrow NaCl(s)}\) |
| R | \(\mathrm{I_2(s)\rightarrow I_2(g)}\) |
| S | \(\mathrm{2NO_2(g)\rightarrow N_2O_4(g)}\) |
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: R
Explanation: Sublimation of iodine, \(\mathrm{I_2(s)\rightarrow I_2(g)}\), changes an ordered solid into a gas. This greatly increases molecular freedom and accessible arrangements, so \(\Delta S_{\text{system}} \gt 0\). Condensation of steam to liquid water decreases entropy. Crystallization from aqueous ions to solid \(\mathrm{NaCl(s)}\) also generally decreases disorder. The dimerization \(\mathrm{2NO_2(g)\rightarrow N_2O_4(g)}\) decreases the number of gas moles and tends to decrease entropy. The phase change to gas is the strongest positive entropy clue in the table.
311. A spontaneous process is best described as a process that
ⓐ. must occur instantly
ⓑ. must always release heat
ⓒ. must always have \(\Delta H=0\)
ⓓ. has a natural tendency to occur under given conditions
Correct Answer: has a natural tendency to occur under given conditions
Explanation: A spontaneous process has a natural tendency to occur under the specified conditions. Spontaneous does not mean fast; a process may be thermodynamically favored but kinetically slow. It also does not mean that heat must always be released. Some spontaneous processes are endothermic because entropy and temperature effects can favor them. The sign of \(\Delta H\) alone is not enough to decide spontaneity in general. A complete criterion often uses entropy of the universe or Gibbs energy under constant temperature and pressure.
312. The second law of thermodynamics states that for a spontaneous process, the entropy change of the universe is
ⓐ. positive
ⓑ. negative
ⓒ. always zero
ⓓ. equal to pressure
Correct Answer: positive
Explanation: The second law states that the entropy of the universe increases for a spontaneous process. This is written as \(\Delta S_{\text{universe}} \gt 0\). For a reversible process, the entropy change of the universe is zero. The universe here means system plus surroundings. The entropy change of the system alone may be positive or negative, but the total for system and surroundings decides spontaneity. The criterion is about entropy, not pressure.
313. The relation connecting entropy changes of system, surroundings, and universe is
ⓐ. \(\Delta S_{\text{universe}}=\Delta S_{\text{system}}+\Delta S_{\text{surroundings}}\)
ⓑ. \(\Delta S_{\text{universe}}=\Delta S_{\text{system}}-\Delta S_{\text{surroundings}}\)
ⓒ. \(\Delta S_{\text{universe}}=q+w\)
ⓓ. \(\Delta S_{\text{universe}}=\frac{\Delta H}{\Delta U}\)
Correct Answer: \(\Delta S_{\text{universe}}=\Delta S_{\text{system}}+\Delta S_{\text{surroundings}}\)
Explanation: The thermodynamic universe is made of the system and surroundings. Therefore, the total entropy change is the sum of the entropy changes of these two parts. The relation is \(\Delta S_{\text{universe}}=\Delta S_{\text{system}}+\Delta S_{\text{surroundings}}\). A spontaneous process requires this sum to be positive. The system entropy alone cannot always decide spontaneity because the surroundings may gain or lose entropy too. This is why heat flow into or out of surroundings is important in entropy reasoning.
314. At constant temperature and pressure, the entropy change of the surroundings for heat exchange with a system is commonly written as
ⓐ. \(\Delta S_{\text{surroundings}}=+\frac{\Delta H_{\text{system}}}{T}\)
ⓑ. \(\Delta S_{\text{surroundings}}=-\frac{\Delta H_{\text{system}}}{T}\)
ⓒ. \(\Delta S_{\text{surroundings}}=\Delta H_{\text{system}}T\)
ⓓ. \(\Delta S_{\text{surroundings}}=0\) for every reaction
Correct Answer: \(\Delta S_{\text{surroundings}}=-\frac{\Delta H_{\text{system}}}{T}\)
Explanation: At constant pressure, heat exchanged by the system is related to \(\Delta H_{\text{system}}\). If the system releases heat, the surroundings receive it; if the system absorbs heat, the surroundings lose it. Therefore, the surroundings heat is opposite in sign to the system enthalpy change. This gives \(\Delta S_{\text{surroundings}}=-\frac{\Delta H_{\text{system}}}{T}\). The temperature must be in \(\text{K}\). The negative sign is essential because it connects the system's heat change to the surroundings' opposite heat change.
315. A reaction at \(300\,\text{K}\) has \(\Delta H_{\text{system}}=-30.0\,\text{kJ mol}^{-1}\). The entropy change of the surroundings is
ⓐ. \(-100\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓑ. \(+100\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓒ. \(+0.100\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓓ. \(-9000\,\text{J mol}^{-1}\text{K}^{-1}\)
Correct Answer: \(+100\,\text{J mol}^{-1}\text{K}^{-1}\)
Explanation: \( \textbf{Given data:} \) \(\Delta H_{\text{system}}=-30.0\,\text{kJ mol}^{-1}\), \(T=300\,\text{K}\).
\( \textbf{Convert enthalpy to joules:} \)
\[
-30.0\,\text{kJ mol}^{-1}=-30000\,\text{J mol}^{-1}
\]
\( \textbf{Surroundings entropy relation:} \)
\[
\Delta S_{\text{surroundings}}=-\frac{\Delta H_{\text{system}}}{T}
\]
\( \textbf{Substitution:} \)
\[
\Delta S_{\text{surroundings}}=-\frac{-30000\,\text{J mol}^{-1}}{300\,\text{K}}
\]
\( \textbf{Calculation:} \)
\[
\Delta S_{\text{surroundings}}=+100\,\text{J mol}^{-1}\text{K}^{-1}
\]
\( \textbf{Final answer:} \) \(\Delta S_{\text{surroundings}}=+100\,\text{J mol}^{-1}\text{K}^{-1}\). An exothermic system increases the entropy of the surroundings by giving heat to them.
316. A process has \(\Delta S_{\text{system}}=+40\,\text{J K}^{-1}\) and \(\Delta S_{\text{surroundings}}=-25\,\text{J K}^{-1}\). The process is
ⓐ. spontaneous because \(\Delta S_{\text{universe}} \gt 0\)
ⓑ. non-spontaneous because \(\Delta S_{\text{system}}\) is positive
ⓒ. reversible because both entropy changes are nonzero
ⓓ. impossible because surroundings entropy is negative
Correct Answer: spontaneous because \(\Delta S_{\text{universe}} \gt 0\)
Explanation: \( \textbf{System entropy change:} \) \(\Delta S_{\text{system}}=+40\,\text{J K}^{-1}\).
\( \textbf{Surroundings entropy change:} \) \(\Delta S_{\text{surroundings}}=-25\,\text{J K}^{-1}\).
\( \textbf{Total entropy relation:} \)
\[
\Delta S_{\text{universe}}=\Delta S_{\text{system}}+\Delta S_{\text{surroundings}}
\]
\( \textbf{Substitution:} \)
\[
\Delta S_{\text{universe}}=+40\,\text{J K}^{-1}+(-25\,\text{J K}^{-1})
\]
\( \textbf{Calculation:} \)
\[
\Delta S_{\text{universe}}=+15\,\text{J K}^{-1}
\]
\( \textbf{Decision:} \) Since \(\Delta S_{\text{universe}} \gt 0\), the process is spontaneous.
\( \textbf{Final answer:} \) The process is spontaneous. A negative surroundings entropy change can be outweighed by a larger positive system entropy change.
317. Gibbs energy \(G\) is related to enthalpy and entropy by
ⓐ. \(G=H+TS\)
ⓑ. \(G=U+p\)
ⓒ. \(G=H-TS\)
ⓓ. \(G=q+w\)
Correct Answer: \(G=H-TS\)
Explanation: Gibbs energy is defined as \(G=H-TS\). Here \(H\) is enthalpy, \(T\) is absolute temperature, and \(S\) is entropy. The relation combines energy content and entropy effect into one state function. Gibbs energy is especially useful for predicting spontaneity at constant temperature and pressure. It is not equal to heat plus work. The temperature must be in \(\text{K}\) when using Gibbs energy relations.
318. At constant temperature and pressure, the condition for a spontaneous process is
ⓐ. \(\Delta G \lt 0\)
ⓑ. \(\Delta G \gt 0\)
ⓒ. \(\Delta G=0\) only
ⓓ. \(\Delta H=0\) always
Correct Answer: \(\Delta G \lt 0\)
Explanation: Gibbs energy gives a convenient spontaneity criterion at constant temperature and pressure. If \(\Delta G \lt 0\), the process is spontaneous in the forward direction under those conditions. If \(\Delta G \gt 0\), the forward process is non-spontaneous. If \(\Delta G=0\), the system is at equilibrium with respect to that process. This criterion includes both enthalpy and entropy effects through \(\Delta G=\Delta H-T\Delta S\). It is more complete than judging by \(\Delta H\) alone.
319. The change in Gibbs energy at constant temperature is given by
ⓐ. \(\Delta G=\Delta H+T\Delta S\)
ⓑ. \(\Delta G=\Delta H-T\Delta S\)
ⓒ. \(\Delta G=\Delta U+q\)
ⓓ. \(\Delta G=T-\Delta S\)
Correct Answer: \(\Delta G=\Delta H-T\Delta S\)
Explanation: From \(G=H-TS\), at constant temperature the change is written as \(\Delta G=\Delta H-T\Delta S\). This equation combines the enthalpy change and entropy change of the system. The term \(T\Delta S\) has energy units when \(T\) is in \(\text{K}\) and \(\Delta S\) is in \(\text{J K}^{-1}\) or \(\text{J mol}^{-1}\text{K}^{-1}\). A common calculation error is to use \(\Delta S\) in \(\text{J}\) while \(\Delta H\) is in \(\text{kJ}\) without conversion. The minus sign before \(T\Delta S\) is central to deciding temperature effects.
320. A reaction has \(\Delta H=+40.0\,\text{kJ mol}^{-1}\) and \(\Delta S=+120\,\text{J mol}^{-1}\text{K}^{-1}\) at \(300\,\text{K}\). The value of \(\Delta G\) is
ⓐ. \(-4.0\,\text{kJ mol}^{-1}\)
ⓑ. \(+76.0\,\text{kJ mol}^{-1}\)
ⓒ. \(-76.0\,\text{kJ mol}^{-1}\)
ⓓ. \(+4.0\,\text{kJ mol}^{-1}\)
Correct Answer: \(+4.0\,\text{kJ mol}^{-1}\)
Explanation: \( \textbf{Given data:} \) \(\Delta H=+40.0\,\text{kJ mol}^{-1}\), \(\Delta S=+120\,\text{J mol}^{-1}\text{K}^{-1}\), \(T=300\,\text{K}\).
\( \textbf{Convert entropy to \(\text{kJ mol}^{-1}\text{K}^{-1}\):} \)
\[
120\,\text{J mol}^{-1}\text{K}^{-1}=0.120\,\text{kJ mol}^{-1}\text{K}^{-1}
\]
\( \textbf{Gibbs relation:} \)
\[
\Delta G=\Delta H-T\Delta S
\]
\( \textbf{Calculate entropy term:} \)
\[
T\Delta S=(300\,\text{K})(0.120\,\text{kJ mol}^{-1}\text{K}^{-1})
\]
\[
T\Delta S=36.0\,\text{kJ mol}^{-1}
\]
\( \textbf{Substitute:} \)
\[
\Delta G=40.0\,\text{kJ mol}^{-1}-36.0\,\text{kJ mol}^{-1}
\]
\[
\Delta G=+4.0\,\text{kJ mol}^{-1}
\]
\( \textbf{Final answer:} \) \(\Delta G=+4.0\,\text{kJ mol}^{-1}\). At \(300\,\text{K}\), the positive entropy term is not large enough to overcome the positive enthalpy term.
