401. For the same substance under comparable conditions, entropy is generally greatest in the gaseous state because
ⓐ. entropy is identical with enthalpy
ⓑ. gas particles have no possible arrangements
ⓒ. gas particles have greater freedom of movement and accessible arrangements
ⓓ. entropy must be zero for every gas
Correct Answer: gas particles have greater freedom of movement and accessible arrangements
Explanation: Entropy is related to the dispersal of energy and the number of accessible arrangements available to a system. In the gaseous state, particles are much farther apart and can occupy many more positions than in a liquid or solid. This usually gives gases greater entropy than liquids and solids of the same substance under comparable conditions. Entropy is not identical with enthalpy; the two are different state functions with different units. A gas does not have zero entropy simply because it is a gas. The key idea is that greater molecular freedom generally means greater entropy.
402. A solid sample melts at its melting point to form a liquid. The entropy change of the sample is best linked with
ⓐ. an increase in accessible molecular arrangements
ⓑ. complete stoppage of molecular motion
ⓒ. conversion of heat into atomic number
ⓓ. a decrease in freedom of movement
Correct Answer: an increase in accessible molecular arrangements
Explanation: Melting changes a solid into a liquid while the chemical identity of the substance remains the same. In the liquid state, particles have more freedom of movement than in the ordered solid structure. This gives the system more accessible arrangements, so the entropy of the sample increases. The particles do not stop moving during melting. The process also does not change atomic numbers or convert heat into matter. Entropy reasoning should focus on molecular freedom and dispersal, not on a change in chemical formula.
403. When two different gases mix spontaneously at the same temperature, the entropy of the gas system usually increases because
ⓐ. entropy is defined as negative for every gas mixture
ⓑ. enthalpy and entropy become identical during mixing
ⓒ. the particles gain more accessible spatial arrangements
ⓓ. mixing always removes all molecular motion
Correct Answer: the particles gain more accessible spatial arrangements
Explanation: Mixing gives gas particles access to a larger set of possible positions and arrangements. The chemical identities of the gases need not change for entropy to increase. The increase is connected with dispersal and molecular freedom, not with making entropy negative. Enthalpy and entropy remain different thermodynamic functions with different units. Molecular motion is not removed during mixing; rather, the number of accessible arrangements increases. This makes gas mixing a common example of a process with positive entropy change for the system.
404. An element in its standard state has \(\Delta_fH^\circ=0\), but this does not mean its entropy must be zero. The best reason is that
ⓐ. formation-enthalpy zero is a separate convention
ⓑ. all elemental gases have zero entropy
ⓒ. entropy and enthalpy must always have the same reference value
ⓓ. oxygen gas cannot have molecular motion
Correct Answer: formation-enthalpy zero is a separate convention
Explanation: The value \(\Delta_fH^\circ=0\) for an element in its standard state is a convention used for standard enthalpy of formation. Entropy is a different thermodynamic property and is related to accessible arrangements and energy dispersal. A substance such as \(\mathrm{O_2(g)}\) at ordinary conditions has molecular motion and accessible arrangements, so its entropy need not be zero. The reference convention for formation enthalpy should not be transferred automatically to entropy. Enthalpy and entropy also have different units. This distinction prevents mixing two separate thermodynamic ideas.
405. Study the table about entropy statements.
| Statement | Evaluation |
|---|
| P. Entropy generally increases from solid to liquid to gas | correct |
| Q. Entropy is a state function | correct |
| R. Molar entropy has a unit containing \(\text{K}^{-1}\) | correct |
| S. Entropy has the same unit as enthalpy | correct |
The row with incorrect evaluation is
ⓐ. P
ⓑ. S
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: Row P is correctly evaluated because gases usually have greater entropy than liquids and solids of the same substance under comparable conditions. Row Q is also correct because entropy is a state function, so its change depends on the initial and final states. Row R is correct because molar entropy is commonly reported in \(\text{J mol}^{-1}\text{K}^{-1}\). Row S has the wrong evaluation because entropy and enthalpy do not have the same units. Enthalpy is commonly reported in \(\text{kJ mol}^{-1}\), while molar entropy includes the extra temperature factor \(\text{K}^{-1}\). The unit difference reflects the different meanings of the two functions.
406. For a reaction, standard entropy change is calculated using
ⓐ. \(\Delta_rS^\circ=\sum \nu S^\circ(\text{products})-\sum \nu S^\circ(\text{reactants})\)
ⓑ. \(\Delta_rS^\circ=\sum \nu S^\circ(\text{reactants})-\sum \nu S^\circ(\text{products})\)
ⓒ. \(\Delta_rS^\circ=\Delta H^\circ-T\Delta G^\circ\)
ⓓ. \(\Delta_rS^\circ=0\) for every balanced reaction
Correct Answer: \(\Delta_rS^\circ=\sum \nu S^\circ(\text{products})-\sum \nu S^\circ(\text{reactants})\)
Explanation: Entropy is a state function, so reaction entropy change is found by comparing final and initial states. For a reaction, products form the final side and reactants form the initial side. Each standard molar entropy must be multiplied by its stoichiometric coefficient. The correct relation is products minus reactants. Reversing the order would give the opposite sign. The entropy change is not zero for every balanced reaction because molecular number, phase, and structure may change.
407. Use the following standard molar entropies:
\(\mathrm{S^\circ[N_2(g)]=192\,J\,mol^{-1}K^{-1}}\),
\(\mathrm{S^\circ[H_2(g)]=131\,J\,mol^{-1}K^{-1}}\),
\(\mathrm{S^\circ[NH_3(g)]=193\,J\,mol^{-1}K^{-1}}\).
For \(\mathrm{N_2(g)+3H_2(g)\rightarrow2NH_3(g)}\), \(\Delta_rS^\circ\) is closest to
ⓐ. \(+199\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓑ. \(-199\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓒ. \(-68\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓓ. \(+68\,\text{J mol}^{-1}\text{K}^{-1}\)
Correct Answer: \(-199\,\text{J mol}^{-1}\text{K}^{-1}\)
Explanation: \( \textbf{Use reaction entropy relation:} \)
\[
\Delta_rS^\circ=\sum \nu S^\circ(\text{products})-\sum \nu S^\circ(\text{reactants})
\]
\( \textbf{Products contribution:} \)
\[
2S^\circ[\mathrm{NH_3(g)}]=2(193)
\]
\[
=386\,\text{J mol}^{-1}\text{K}^{-1}
\]
\( \textbf{Reactants contribution:} \)
\[
S^\circ[\mathrm{N_2(g)}]+3S^\circ[\mathrm{H_2(g)}]
\]
\[
=192+3(131)=192+393=585\,\text{J mol}^{-1}\text{K}^{-1}
\]
\( \textbf{Subtract:} \)
\[
\Delta_rS^\circ=386-585=-199\,\text{J mol}^{-1}\text{K}^{-1}
\]
\( \textbf{Final answer:} \) \(\Delta_rS^\circ=-199\,\text{J mol}^{-1}\text{K}^{-1}\). The negative value agrees with the decrease from \(4\) moles of gas to \(2\) moles of gas.
408. Use the graph description below.
A graph of entropy \(S\) versus temperature \(T\) for a pure substance rises with temperature and shows sudden upward jumps at melting and boiling points.
The upward jumps occur mainly because
ⓐ. entropy must decrease during melting and boiling
ⓑ. enthalpy becomes zero at every phase change
ⓒ. gases have no entropy after boiling
ⓓ. phase changes increase accessible arrangements
Correct Answer: phase changes increase accessible arrangements
Explanation: Entropy generally increases as temperature rises because particles access more energy states. At melting and boiling, the physical state changes and molecular freedom increases significantly. These changes produce upward jumps in entropy. The chemical formula may remain the same, as in \(\mathrm{H_2O(s)\rightarrow H_2O(l)\rightarrow H_2O(g)}\). Melting and boiling do not require entropy to decrease. The graph reflects both heating and phase-change contributions to entropy.
409. During a reversible phase change at transition temperature \(T_{\text{tr}}\), the entropy change is related to enthalpy of transition by
ⓐ. \(\Delta S_{\text{tr}}=\Delta H_{\text{tr}}T_{\text{tr}}\)
ⓑ. \(\Delta S_{\text{tr}}=\frac{\Delta H_{\text{tr}}}{T_{\text{tr}}}\)
ⓒ. \(\Delta S_{\text{tr}}=T_{\text{tr}}-\Delta H_{\text{tr}}\)
ⓓ. \(\Delta S_{\text{tr}}=\frac{T_{\text{tr}}}{\Delta H_{\text{tr}}}\)
Correct Answer: \(\Delta S_{\text{tr}}=\frac{\Delta H_{\text{tr}}}{T_{\text{tr}}}\)
Explanation: For a reversible phase transition at its transition temperature, the reversible heat is equal to the enthalpy change of transition. Since entropy change is \(\Delta S=\frac{q_{\text{rev}}}{T}\), the relation becomes \(\Delta S_{\text{tr}}=\frac{\Delta H_{\text{tr}}}{T_{\text{tr}}}\). The temperature must be in \(\text{K}\). Fusion and vaporization entropies can be calculated this way at the melting and boiling points. Multiplying by temperature would give the wrong units. The relation connects phase-change heat with the entropy gained during the transition.
410. The enthalpy of fusion of a substance is \(6.0\,\text{kJ mol}^{-1}\), and its melting point is \(300\,\text{K}\). The entropy of fusion is
ⓐ. \(0.020\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓑ. \(1800\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓒ. \(50\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓓ. \(20\,\text{J mol}^{-1}\text{K}^{-1}\)
Correct Answer: \(20\,\text{J mol}^{-1}\text{K}^{-1}\)
Explanation: \( \textbf{Given data:} \) \(\Delta H_{\text{fus}}=6.0\,\text{kJ mol}^{-1}\), \(T_{\text{m}}=300\,\text{K}\).
\( \textbf{Convert enthalpy:} \)
\[
6.0\,\text{kJ mol}^{-1}=6000\,\text{J mol}^{-1}
\]
\( \textbf{Use phase-change entropy relation:} \)
\[
\Delta S_{\text{fus}}=\frac{\Delta H_{\text{fus}}}{T_{\text{m}}}
\]
\( \textbf{Substitute:} \)
\[
\Delta S_{\text{fus}}=\frac{6000\,\text{J mol}^{-1}}{300\,\text{K}}
\]
\( \textbf{Calculate:} \)
\[
\Delta S_{\text{fus}}=20\,\text{J mol}^{-1}\text{K}^{-1}
\]
\( \textbf{Final answer:} \) \(\Delta S_{\text{fus}}=20\,\text{J mol}^{-1}\text{K}^{-1}\). The value is positive because melting increases molecular freedom.
411. A liquid vaporizes reversibly at its boiling point. If \(\Delta H_{\text{vap}}=40.0\,\text{kJ mol}^{-1}\) and \(T_b=400\,\text{K}\), the entropy of vaporization is
ⓐ. \(-100\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓑ. \(+10\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓒ. \(+100\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓓ. \(+16000\,\text{J mol}^{-1}\text{K}^{-1}\)
Correct Answer: \(+100\,\text{J mol}^{-1}\text{K}^{-1}\)
Explanation: \( \textbf{Given transition:} \) Vaporization changes liquid to gas.
\( \textbf{Given enthalpy:} \)
\[
\Delta H_{\text{vap}}=40.0\,\text{kJ mol}^{-1}=40000\,\text{J mol}^{-1}
\]
\( \textbf{Boiling point:} \)
\[
T_b=400\,\text{K}
\]
\( \textbf{Relation:} \)
\[
\Delta S_{\text{vap}}=\frac{\Delta H_{\text{vap}}}{T_b}
\]
\( \textbf{Substitution:} \)
\[
\Delta S_{\text{vap}}=\frac{40000}{400}
\]
\[
\Delta S_{\text{vap}}=100\,\text{J mol}^{-1}\text{K}^{-1}
\]
\( \textbf{Final answer:} \) \(\Delta S_{\text{vap}}=+100\,\text{J mol}^{-1}\text{K}^{-1}\). Vaporization gives a large positive entropy change because gas has far greater molecular freedom than liquid.
412. A reaction has \(\Delta H^\circ=-80\,\text{kJ mol}^{-1}\) and \(\Delta S^\circ=-160\,\text{J mol}^{-1}\text{K}^{-1}\). The temperature at which \(\Delta G^\circ=0\) is
ⓐ. \(200\,\text{K}\)
ⓑ. \(500\,\text{K}\)
ⓒ. \(800\,\text{K}\)
ⓓ. \(12800\,\text{K}\)
Correct Answer: \(500\,\text{K}\)
Explanation: \( \textbf{Threshold condition:} \) \(\Delta G^\circ=0\).
\( \textbf{Gibbs relation:} \)
\[
\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ
\]
\( \textbf{At threshold:} \)
\[
0=\Delta H^\circ-T\Delta S^\circ
\]
\[
T=\frac{\Delta H^\circ}{\Delta S^\circ}
\]
\( \textbf{Convert enthalpy to joules:} \)
\[
-80\,\text{kJ mol}^{-1}=-80000\,\text{J mol}^{-1}
\]
\( \textbf{Substitute signs:} \)
\[
T=\frac{-80000\,\text{J mol}^{-1}}{-160\,\text{J mol}^{-1}\text{K}^{-1}}
\]
\[
T=500\,\text{K}
\]
\( \textbf{Temperature interpretation:} \) With both \(\Delta H^\circ\) and \(\Delta S^\circ\) negative, the reaction is favoured below this temperature.
\( \textbf{Final answer:} \) \(T=500\,\text{K}\). Above this value, the entropy term opposes the reaction strongly enough to make \(\Delta G^\circ\) positive.
413. A reaction has the following standard data at \(298\,\text{K}\):
\(\Delta H^\circ=-120\,\text{kJ mol}^{-1}\) and \(\Delta S^\circ=-300\,\text{J mol}^{-1}\text{K}^{-1}\).
Its \(\Delta G^\circ\) and spontaneity under standard conditions are closest to
ⓐ. \(-209.4\,\text{kJ mol}^{-1}\), spontaneous
ⓑ. \(+30.6\,\text{kJ mol}^{-1}\), non-spontaneous
ⓒ. \(+209.4\,\text{kJ mol}^{-1}\), non-spontaneous
ⓓ. \(-30.6\,\text{kJ mol}^{-1}\), spontaneous
Correct Answer: \(-30.6\,\text{kJ mol}^{-1}\), spontaneous
Explanation: \( \textbf{Given data:} \) \(\Delta H^\circ=-120\,\text{kJ mol}^{-1}\), \(\Delta S^\circ=-300\,\text{J mol}^{-1}\text{K}^{-1}\), \(T=298\,\text{K}\).
\( \textbf{Convert entropy:} \)
\[
-300\,\text{J mol}^{-1}\text{K}^{-1}=-0.300\,\text{kJ mol}^{-1}\text{K}^{-1}
\]
\( \textbf{Use Gibbs relation:} \)
\[
\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ
\]
\( \textbf{Entropy term:} \)
\[
T\Delta S^\circ=(298)(-0.300)=-89.4\,\text{kJ mol}^{-1}
\]
\( \textbf{Substitution:} \)
\[
\Delta G^\circ=-120-(-89.4)
\]
\[
\Delta G^\circ=-30.6\,\text{kJ mol}^{-1}
\]
\( \textbf{Decision:} \) \(\Delta G^\circ \lt 0\), so the reaction is spontaneous under standard conditions at \(298\,\text{K}\).
\( \textbf{Final answer:} \) \(\Delta G^\circ\approx-30.6\,\text{kJ mol}^{-1}\), spontaneous. The negative entropy change weakens but does not remove the favourable enthalpy effect at this temperature.
414. The following signs are reported for four reactions.
| Reaction | \(\Delta H^\circ\) | \(\Delta S^\circ\) | Temperature favourability |
|---|
| P | \(-\) | \(+\) | all \(T\) |
| Q | \(+\) | \(+\) | high \(T\) |
| R | \(-\) | \(-\) | low \(T\) |
| S | \(+\) | \(-\) | all \(T\) |
The row with the wrong temperature favourability is
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: For P, \(\Delta H^\circ \lt 0\) and \(\Delta S^\circ \gt 0\), both terms make \(\Delta G^\circ\) negative, so all temperatures favour it. For Q, positive entropy can overcome positive enthalpy at high temperature. For R, negative enthalpy favours the reaction, but negative entropy opposes it more strongly at high temperature, so low temperature favours it. For S, \(\Delta H^\circ \gt 0\) and \(\Delta S^\circ \lt 0\), both terms make \(\Delta G^\circ\) positive. Therefore, S is not favoured at any positive temperature. The sign combination, not the table label, decides the result.
415. A reaction has \(\Delta G^\circ=-5.0\,\text{kJ mol}^{-1}\) at \(300\,\text{K}\). If the reaction is multiplied by \(2\), the corresponding equilibrium constant becomes
ⓐ. \(\sqrt{K}\)
ⓑ. \(-K\)
ⓒ. \(K^2\)
ⓓ. \(\frac{1}{K}\)
Correct Answer: \(K^2\)
Explanation: Multiplying a thermochemical equation by \(2\) doubles \(\Delta G^\circ\). The equilibrium expression also changes because every stoichiometric coefficient becomes doubled, so every concentration or pressure factor gets doubled power. Therefore, the new equilibrium constant is \(K^2\). This is consistent with \(\Delta G^\circ=-RT\ln K\), because doubling \(\Delta G^\circ\) doubles \(\ln K\). If \(\ln K\) doubles, \(K\) becomes squared. The reciprocal \(\frac{1}{K}\) would correspond to reversing the reaction, not multiplying it by \(2\).
416. A reaction has \(K=9\) at a fixed temperature. If the same reaction equation is divided by \(2\), the new equilibrium constant is
ⓐ. \(9\)
ⓑ. \(81\)
ⓒ. \(\frac{1}{9}\)
ⓓ. \(3\)
Correct Answer: \(3\)
Explanation: Dividing a balanced reaction by \(2\) halves all stoichiometric coefficients. In the equilibrium expression, this halves all exponents. Therefore, the new equilibrium constant becomes the square root of the original value. For \(K=9\), the new value is \(\sqrt{9}=3\). The value is not unchanged because the equation as written has changed. The reciprocal would be used if the reaction were reversed. Equilibrium constants must always be interpreted for the exact reaction equation written.
417. A student writes, “Because \(\Delta G^\circ=-RT\ln K\), a larger \(K\) always makes \(\Delta G^\circ\) more positive.” The correction is that
ⓐ. \(K\) increase makes \(\Delta G^\circ\) more negative
ⓑ. \(K\) has no logarithm in the relation
ⓒ. \(R\) is negative, so the sign changes
ⓓ. \(\Delta G^\circ\) is positive for every value of \(K\)
Correct Answer: \(K\) increase makes \(\Delta G^\circ\) more negative
Explanation: The relation is \(\Delta G^\circ=-RT\ln K\). Since \(R\) and \(T\) are positive, the sign of \(\Delta G^\circ\) is controlled by the negative sign and the value of \(\ln K\). If \(K\) increases above \(1\), \(\ln K\) becomes increasingly positive. Multiplication by \(-RT\) then makes \(\Delta G^\circ\) increasingly negative. This means products are more strongly favoured under standard-state comparison. The negative sign in the formula must not be overlooked.
418. A reaction mixture has \(Q=0.01\), and the equilibrium constant is \(K=100\) at the same temperature. The forward reaction is strongly favoured because
ⓐ. \(Q\) is much smaller than \(K\)
ⓑ. \(Q\) is smaller than \(1\), so equilibrium is impossible
ⓒ. \(K\) must be converted into \(\text{kJ mol}^{-1}\)
ⓓ. \(Q\) and \(K\) must be added before comparison
Correct Answer: \(Q\) is much smaller than \(K\)
Explanation: Direction is predicted by comparing \(Q\) and \(K\). Here \(Q=0.01\), while \(K=100\), so \(Q\ll K\). The mixture has far less product relative to reactant than the equilibrium mixture would have. The forward reaction is therefore favoured to increase \(Q\). Equilibrium is not impossible just because \(Q \lt 1\); it depends on whether \(Q\) equals \(K\). Reaction quotient and equilibrium constant are dimensionless composition ratios in the thermodynamic expression, not energy values to be converted into \(\text{kJ mol}^{-1}\).
419. A reaction has \(\Delta G=0\) at its current composition but \(\Delta G^\circ=+7.5\,\text{kJ mol}^{-1}\). The equilibrium constant at that temperature must be
ⓐ. equal to \(1\)
ⓑ. less than \(1\)
ⓒ. greater than \(1\)
ⓓ. negative
Correct Answer: less than \(1\)
Explanation: The current condition \(\Delta G=0\) means the mixture is at equilibrium, so \(Q=K\). The sign of \(\Delta G^\circ\) is related to \(K\) by \(\Delta G^\circ=-RT\ln K\). If \(\Delta G^\circ\) is positive, then \(\ln K\) must be negative. A negative logarithm means \(K \lt 1\). The actual mixture can still be at equilibrium even when \(\Delta G^\circ\) is not zero. The positive standard Gibbs energy shows that reactants are favoured under the standard-state comparison.
420. A final comparison of criteria is given below.
| Criterion | Meaning under suitable conditions |
|---|
| P. \(\Delta S_{\text{universe}} \gt 0\) | spontaneous process |
| Q. \(\Delta G \lt 0\) at constant \(T,p\) | forward process favoured |
| R. \(\Delta G=0\) at constant \(T,p\) | equilibrium |
| S. \(Q=K\) | equilibrium composition |
The correctly stated criteria are
ⓐ. P and Q only
ⓑ. Q and R only
ⓒ. P, Q, R, and S
ⓓ. P, Q, and R only
Correct Answer: P, Q, R, and S
Explanation: The entropy criterion says that a spontaneous process has \(\Delta S_{\text{universe}} \gt 0\). At constant temperature and pressure, the Gibbs criterion gives \(\Delta G \lt 0\) for a forward spontaneous process. When \(\Delta G=0\), the system is at equilibrium under those conditions. The composition form of equilibrium is \(Q=K\). All four statements are consistent ways of describing thermodynamic direction or equilibrium, each with its own condition. The table connects entropy, Gibbs energy, and reaction quotient language into one final framework.
