Ap Chemistry 2차 - Unit 6

Unit 6: Thermodynamics — Multiple Choice Questions

Exothermic vs Endothermic Calorimetry (Q = mcΔT) Calorimeter Errors (HHI) Hess’s Law Bond & Formation Enthalpy Entropy (ΔS°) Gibbs Free Energy (ΔG°) Equilibrium (K) Electrochem Link (ΔG° = −nFE°)

1. An exothermic reaction has which sign for ΔH (system)?

(A) ΔH > 0
(B) ΔH < 0
(C) ΔH = 0
(D) Sign cannot be defined

View Answer

Correct Answer: (B) Explanation: Exothermic releases heat to surroundings ⇒ system enthalpy decreases.

2. A solution warms up during a reaction at constant pressure. The sign of \(q_{\text{rxn}}\) (system) is:

(A) Positive
(B) Negative
(C) Zero
(D) Depends on volume

View Answer

Correct Answer: (B) Explanation: Solution (surroundings) absorbs heat ⇒ \(q_{\text{rxn}} = -q_{\text{soln}} < 0\).

3. In a coffee-cup calorimeter (constant pressure), the measured heat corresponds most directly to:

(A) ΔU of reaction
(B) ΔH of reaction
(C) Work only
(D) ΔG of reaction

View Answer

Correct Answer: (B) Explanation: At constant pressure with negligible PV work other than expansion, \(q_p \approx \Delta H_{\text{rxn}}\).

4. 100.0 g of water (\(c=4.184\ \text{J·g}^{-1}\text{·K}^{-1}\)) warms by 2.50 K during a dissolution. If 0.0500 mol of solute reacted, estimate \(\Delta H_{\text{rxn}}\) (per mol, sign included).

(A) \(+21.0\ \text{kJ·mol}^{-1}\)
(B) \(-21.0\ \text{kJ·mol}^{-1}\)
(C) \(-10.5\ \text{kJ·mol}^{-1}\)
(D) \(+10.5\ \text{kJ·mol}^{-1}\)

View Answer

Correct Answer: (B) Explanation: \(q_{\text{soln}}=mc\Delta T=100\times4.184\times2.50\approx 1.046\ \text{kJ}\). System lost that heat: \(q_{\text{rxn}}\approx -1.046\ \text{kJ}\). Per 0.0500 mol: \(\Delta H\approx -1.046/0.0500\approx -21.0\ \text{kJ·mol}^{-1}\).

5. Which source of error decreases the magnitude of measured \(|\Delta H|\) for an exothermic reaction in solution?

(A) Heat loss to surroundings (H)
(B) Heat absorbed by the calorimeter/equipment (H)
(C) Incomplete reaction/combustion (I)
(D) All of the above (HHI)

View Answer

Correct Answer: (D) Explanation: All reduce observed temperature rise ⇒ \(|\Delta H|\) appears smaller.

6. A bomb calorimeter has \(C_{\text{cal}}=6.20\ \text{kJ·K}^{-1}\). Combustion raises temperature by 1.35 K for a 0.500 g sample (M = 78.0 g·mol\(^{-1}\)). Find \(\Delta U_{\text{comb}}\) (kJ·mol\(^{-1}\)).

(A) \(-654\)
(B) \(-987\)
(C) \(-1306\)
(D) \(-1750\)

View Answer

Correct Answer: (C) Explanation: \(q_{\text{cal}}=C\Delta T=8.37\ \text{kJ}\Rightarrow q_{\text{rxn}}=-8.37\ \text{kJ}\). Moles \(=0.500/78.0=6.41\times10^{-3}\), so \(\Delta U\approx -8.37/6.41\times10^{-3}\approx -1.31\times10^{3}\ \text{kJ·mol}^{-1}\).

7. Given: \( \text{A}\to\text{B}\) ΔH = +100 kJ; \( \text{B}\to\text{C}\) ΔH = −40 kJ. What is ΔH for \(2\text{A}\to2\text{C}\)?

(A) +60 kJ
(B) +120 kJ
(C) −60 kJ
(D) −120 kJ

View Answer

Correct Answer: (B) Explanation: \(\text{A}\to\text{C} = 100 + (-40) = +60\ \text{kJ}\); doubling ⇒ +120 kJ.

8. Using average bond energies, estimate ΔH for \(\text{H}_2+\text{Cl}_2\to 2\text{HCl}\). (Use kJ·mol\(^{-1}\): D(H–H)=436, D(Cl–Cl)=243, D(H–Cl)=431.)

(A) +183
(B) −183
(C) −86
(D) +86

View Answer

Correct Answer: (B) Explanation: ΔH ≈ bonds broken − formed \(=(436+243)−2×431=679−862=−183\).

9. Compute ΔH° for \(\text{CH}_4(g)+2\text{O}_2(g)\to \text{CO}_2(g)+2\text{H}_2\text{O}(l)\) using ΔHf° (kJ·mol\(^{-1}\)): CH\(_4\)=−74.8, CO\(_2\)=−393.5, H\(_2\)O(l)=−285.8, O\(_2\)=0.

(A) −802
(B) −890
(C) −242
(D) +890

View Answer

Correct Answer: (B) Explanation: Σproducts − Σreactants \(=[-393.5+2(−285.8)]−[−74.8]=−890.3\ \text{kJ·mol}^{-1}\).

10. Which change most likely has ΔS° < 0?

(A) \(\text{N}_2\text{O}_4(g)\to 2\text{NO}_2(g)\)
(B) \(\text{H}_2\text{O}(l)\to \text{H}_2\text{O}(g)\)
(C) \(2\text{NO}_2(g)\to \text{N}_2\text{O}_4(g)\)
(D) NaCl(s) dissolves in water

View Answer

Correct Answer: (C) Explanation: Fewer gas moles lowers disorder ⇒ ΔS° negative.

11. Given \(S^\circ\) (J·mol\(^{-1}\)·K\(^{-1}\)): A(g)=180, B(g)=220. For \(\text{A}(g)\to \text{B}(g)\), ΔS° equals:

(A) −40
(B) 0
(C) +40
(D) +400

View Answer

Correct Answer: (C) Explanation: Σproducts − Σreactants \(=220−180=+40\ \text{J·mol}^{-1}\text{·K}^{-1}\).

12. For ΔH = +50.0 kJ·mol\(^{-1}\) and ΔS = +120 J·mol\(^{-1}\)·K\(^{-1}\), ΔG at 298 K is approximately:

(A) −14.2 kJ·mol\(^{-1}\)
(B) +14.2 kJ·mol\(^{-1}\)
(C) +85.8 kJ·mol\(^{-1}\)
(D) −85.8 kJ·mol\(^{-1}\)

View Answer

Correct Answer: (B) Explanation: Convert ΔS→0.120 kJ·mol\(^{-1}\)·K\(^{-1}\): \(ΔG=50.0−(298)(0.120)=+14.2\ \text{kJ·mol}^{-1}\).

13. With ΔH = +50.0 kJ·mol\(^{-1}\) and ΔS = +120 J·mol\(^{-1}\)·K\(^{-1}\), the reaction becomes spontaneous above approximately:

(A) 298 K
(B) 350 K
(C) 417 K
(D) 600 K

View Answer

Correct Answer: (C) Explanation: Threshold \(T^*=\Delta H/\Delta S=50{,}000/120\approx 417\ \text{K}\).

14. If ΔG° = −8.00 kJ·mol\(^{-1}\) at 298 K, what is \(K\)? (Use \(R=8.314\ \text{J·mol}^{-1}\text{·K}^{-1}\))

(A) 0.040
(B) 0.40
(C) 2.5
(D) 25

View Answer

Correct Answer: (D) Explanation: \(ΔG^\circ=-RT\ln K\Rightarrow \ln K=8000/(8.314×298)\approx 3.23\Rightarrow K\approx e^{3.23}\approx 25\).

15. If ΔG° = +5.70 kJ·mol\(^{-1}\) at 298 K, \(K\) is closest to:

(A) 10
(B) 1.0
(C) 0.10
(D) 0.010

View Answer

Correct Answer: (C) Explanation: \(\ln K= -5700/(8.314×298)\approx -2.30\Rightarrow K\approx e^{-2.30}\approx 0.10\).

16. Which pair is correct at a given T?

(A) \(K=1\) ↔ ΔG° < 0
(B) \(K\gg1\) ↔ ΔG° ≈ 0
(C) \(K=1\) ↔ ΔG° = 0
(D) \(K\ll1\) ↔ ΔG° < 0

View Answer

Correct Answer: (C) Explanation: \(ΔG^\circ=-RT\ln K\). If \(K=1\), \(\ln 1=0\Rightarrow ΔG^\circ=0\).

17. If ΔG° = −40 kJ·mol\(^{-1}\) at 298 K, then \(K\) is:

(A) Much less than 1
(B) About 1
(C) Much greater than 1
(D) Cannot tell

View Answer

Correct Answer: (C) Explanation: Large negative ΔG° ⇒ \(K\) is very large.

18. A redox reaction transfers \(n=2\) electrons and has ΔG° = −212 kJ·mol\(^{-1}\). Find \(E^\circ\). (Use \(F=96485\ \text{C·mol}^{-1}\))

(A) 0.55 V
(B) 0.86 V
(C) 1.10 V
(D) 1.50 V

View Answer

Correct Answer: (C) Explanation: \(ΔG^\circ=-nFE^\circ\Rightarrow E^\circ=|ΔG^\circ|/(nF)=212000/(2×96485)\approx 1.10\ \text{V}\).

19. For a cell with \(E^\circ=0.518\ \text{V}\) and \(n=3\), compute ΔG°. (kJ·mol\(^{-1}\))

(A) −150
(B) −100
(C) −50
(D) +150

View Answer

Correct Answer: (A) Explanation: \(ΔG^\circ=-nFE^\circ=-(3)(96485)(0.518)\approx -1.50×10^5\ \text{J·mol}^{-1}=-150\ \text{kJ·mol}^{-1}\).

20. Which combination guarantees spontaneity at all temperatures?

(A) ΔH < 0, ΔS > 0
(B) ΔH < 0, ΔS < 0
(C) ΔH > 0, ΔS > 0
(D) ΔH > 0, ΔS < 0

View Answer

Correct Answer: (A) Explanation: \(ΔG=ΔH−TΔS\) is negative for all \(T\) if exothermic and entropy-favored.

21. A reaction has ΔH = −60.0 kJ·mol\(^{-1}\), ΔS = −50 J·mol\(^{-1}\)·K\(^{-1}\). It is nonspontaneous when \(T\) is:

(A) Below 1200 K
(B) Above 1200 K
(C) At 1200 K only
(D) Never

View Answer

Correct Answer: (B) Explanation: \(T^*=ΔH/ΔS=(-60{,}000)/(-50)=1200\ \text{K}\). For \(T>T^*\), \(ΔG>0\).

22. Which statement about standard formation enthalpies is TRUE?

(A) ΔHf°[O\(_2\)(g)] = −285.8 kJ·mol\(^{-1}\)
(B) ΔHf° equals zero for any compound at 298 K
(C) Elements in their standard states have ΔHf° = 0
(D) ΔHf° is positive by definition

View Answer

Correct Answer: (C) Explanation: Pure elements in their standard state (e.g., O\(_2\)(g)) have ΔHf° = 0.

23. In a constant-pressure calorimetry experiment, the solution temperature drops. What can be concluded about ΔH of the process (system)?

(A) ΔH < 0 (exothermic)
(B) ΔH > 0 (endothermic)
(C) ΔH = 0
(D) Insufficient information

View Answer

Correct Answer: (B) Explanation: Solution (surroundings) cools ⇒ system absorbed heat ⇒ endothermic.

24. Ammonium nitrate dissolves in water with ΔH > 0 yet occurs spontaneously at room temperature. The best explanation is:

(A) ΔS < 0 but small
(B) ΔS > 0 and \(TΔS\) > ΔH
(C) ΔG depends only on ΔH
(D) ΔH, ΔS are irrelevant

View Answer

Correct Answer: (B) Explanation: Increased dispersal upon dissolution gives positive ΔS; \(ΔG=ΔH−TΔS<0\).

25. At the normal melting point of a pure substance under 1 bar, which is TRUE for the phase change (solid ⇌ liquid)?

(A) ΔG = 0
(B) ΔH = 0
(C) ΔS = 0
(D) ΔG° ≠ 0

View Answer

Correct Answer: (A) Explanation: At equilibrium of the two phases, \(ΔG=0\) for the transition at that T,P.

26. Given: A → B ΔH = −50 kJ; B → C ΔH = +70 kJ. What is ΔH for C → A?

(A) −120 kJ
(B) −20 kJ
(C) +20 kJ
(D) +120 kJ

View Answer

Correct Answer: (B) Explanation: A→C = (−50) + (+70) = +20 kJ; reversing gives C→A = −20 kJ.

27. In a coffee-cup calorimetry experiment for an exothermic process, you mistakenly neglect the calorimeter heat capacity (assume C_cal = 0). Your calculated molar ΔH will be:

(A) Too negative (more exothermic than true)
(B) Too positive (less exothermic than true)
(C) Unchanged
(D) Sign reversed

View Answer

Correct Answer: (B) Explanation: Some heat warms the calorimeter; ignoring it underestimates heat released ⇒ ΔH appears less negative.

28. Using standard formation enthalpies, the correct relation for reaction enthalpy is:

(A) ΔH° = ΣΔHf°(reactants) − ΣΔHf°(products)
(B) ΔH° = ΣΔHf°(products) − ΣΔHf°(reactants)
(C) ΔH° = ΣB.E(products) − ΣB.E(reactants)
(D) ΔH° = 0 for any reaction at 298 K

View Answer

Correct Answer: (B) Explanation: With ΔHf°(elements in std. state) = 0, ΔH° = Σ(products) − Σ(reactants).

29. Which process most likely has the largest positive ΔS°?

(A) \(2\text{SO}_2(g)+\text{O}_2(g)\to 2\text{SO}_3(g)\)
(B) \(\text{H}_2\text{O}(l)\to \text{H}_2\text{O}(g)\)
(C) \(2\text{NO}_2(g)\to \text{N}_2\text{O}_4(g)\)
(D) \(\text{CO}_2(g)\to \text{CO}_2(aq)\)

View Answer

Correct Answer: (B) Explanation: Vaporization greatly increases dispersal ⇒ large positive ΔS°.

30. For ΔH > 0 and ΔS < 0, the reaction is:

(A) Spontaneous at all T
(B) Spontaneous only at low T
(C) Spontaneous only at high T
(D) Nonspontaneous at all T

View Answer

Correct Answer: (D) Explanation: \(ΔG=ΔH−TΔS\) is positive for any T when ΔH>0 and ΔS<0.

31. Which statement is TRUE at temperature T?

(A) If \(Q < K\), then ΔG > 0
(B) If \(Q = K\), then ΔG < 0
(C) If \(Q > K\), then ΔG < 0
(D) If \(Q < K\), then ΔG < 0

View Answer

Correct Answer: (D) Explanation: \(ΔG=RT\ln(Q/K)\). If \(Q<K\), \(ΔG<0\) and reaction proceeds forward.

32. At 298 K, \(ΔG^\circ=-5.00\ \text{kJ·mol}^{-1}\). If \(Q=10.0\), what is \(ΔG\)? (R=8.314 J·mol\(^{-1}\)·K\(^{-1}\))

(A) −10.7 kJ·mol\(^{-1}\)
(B) −5.00 kJ·mol\(^{-1}\)
(C) +0.71 kJ·mol\(^{-1}\)
(D) +5.00 kJ·mol\(^{-1}\)

View Answer

Correct Answer: (C) Explanation: \(ΔG=ΔG^\circ+RT\ln Q=-5.00+ (8.314×298×\ln10)/1000\approx -5.00+5.71=+0.71\ \text{kJ·mol}^{-1}\).

33. For a process at constant T,P and reversible conditions, the maximum non-PV (electrical) work obtainable equals:

(A) \(+ΔH\)
(B) \(−ΔH\)
(C) \(+ΔG\)
(D) \(−ΔG\)

View Answer

Correct Answer: (D) Explanation: \(w_{\text{nonexp,max}}=−ΔG\).

34. Which is TRUE for a spontaneous galvanic cell under standard conditions?

(A) \(E^\circ<0\), \(ΔG^\circ<0\)
(B) \(E^\circ>0\), \(ΔG^\circ<0\)
(C) \(E^\circ>0\), \(ΔG^\circ>0\)
(D) \(E^\circ=0\), \(K=1\), always spontaneous

View Answer

Correct Answer: (B) Explanation: \(ΔG^\circ=−nFE^\circ\). Spontaneous ⇒ \(ΔG^\circ<0\) ⇒ \(E^\circ>0\).

35. At 298 K, a reaction has \(n=2\) and \(E^\circ=0.250\ \text{V}\). Find \(K\). (F=96485 C·mol\(^{-1}\), R=8.314 J·mol\(^{-1}\)·K\(^{-1}\))

(A) \(3.0\times10^{3}\)
(B) \(1.2\times10^{6}\)
(C) \(2.9\times10^{8}\)
(D) \(4.1\times10^{10}\)

View Answer

Correct Answer: (C) Explanation: \(ΔG^\circ=−nFE^\circ=−48.2\ \text{kJ·mol}^{-1}\). \(\ln K=|ΔG^\circ|/(RT)\approx 19.48\Rightarrow K\approx e^{19.48}\approx 2.9\times10^8\).

36. The Third Law of Thermodynamics states that the entropy of a perfect crystal at 0 K is:

(A) Undefined
(B) Zero
(C) One arbitrary unit
(D) Equal to its enthalpy

View Answer

Correct Answer: (B) Explanation: \(S\to 0\) for a perfect crystal as \(T\to 0\ \text{K}\).

37. Mixing two different ideal gases at the same T,P results in ΔS:

(A) Negative
(B) Zero
(C) Positive
(D) Sign cannot be predicted

View Answer

Correct Answer: (C) Explanation: Randomization increases microstates ⇒ \(ΔS_{\text{mix}}>0\).

38. When a gas dissolves into a liquid (at constant T), the sign of ΔS for the system is typically:

(A) Positive, because particles spread out
(B) Negative, because gas molecules lose translational freedom
(C) Zero
(D) Dependent only on ΔH

View Answer

Correct Answer: (B) Explanation: Dissolved gas is more constrained ⇒ entropy decreases.

39. Estimate ΔH for \(\text{H}_2+\tfrac{1}{2}\text{O}_2\to \text{H}_2\text{O}(g)\) using average bond energies (kJ·mol\(^{-1}\)): D(H–H)=436, D(O=O)=498, D(O–H)=463.

(A) −241 kJ·mol\(^{-1}\)
(B) −120 kJ·mol\(^{-1}\)
(C) +120 kJ·mol\(^{-1}\)
(D) +241 kJ·mol\(^{-1}\)

View Answer

Correct Answer: (A) Explanation: Broken: 1×436 + 0.5×498 = 685 kJ; formed: 2×463 = 926 kJ; ΔH ≈ 685 − 926 = −241 kJ·mol\(^{-1}\).

40. A reaction in solution warms 200.0 g of solution (c = 4.18 J·g\(^{-1}\)·K\(^{-1}\)) and a calorimeter with \(C_{\text{cal}}=85.0\ \text{J·K}^{-1}\) by 1.20 K. If 0.0100 mol reacted, estimate ΔH (per mol, sign included).

(A) \(+110.5\ \text{kJ·mol}^{-1}\)
(B) \(-110.5\ \text{kJ·mol}^{-1}\)
(C) \(-9.21\ \text{kJ·mol}^{-1}\)
(D) \(+9.21\ \text{kJ·mol}^{-1}\)

View Answer

Correct Answer: (B) Explanation: \(q_{\text{soln}}=200×4.18×1.20=1003\ \text{J}\), \(q_{\text{cal}}=85×1.20=102\ \text{J}\). Total \(=1105\ \text{J}\) to surroundings ⇒ \(q_{\text{rxn}}=-1105\ \text{J}\). Per 0.0100 mol ⇒ \(-110.5\ \text{kJ·mol}^{-1}\).

41. For an exothermic reaction (ΔH° < 0), increasing temperature generally causes the equilibrium constant \(K\) to:

(A) Increase
(B) Decrease
(C) Stay exactly the same
(D) Oscillate with T

View Answer

Correct Answer: (B) Explanation: Higher T disfavors heat-releasing direction ⇒ \(K\) decreases (consistent with \(ΔG^\circ=ΔH^\circ−TΔS^\circ\)).

42. Which species has the largest standard molar entropy at 298 K?

(A) \(\text{C}_2\text{H}_2(g)\)
(B) \(\text{C}_2\text{H}_6(g)\)
(C) \(\text{CH}_4(g)\)
(D) \(\text{C}_2\text{H}_6(l)\)

View Answer

Correct Answer: (B) Explanation: Greater molecular complexity and gas phase increase S°; among gases, ethane > methane > acetylene typically.

43. For \(\text{H}_2\text{O}(s)\to \text{H}_2\text{O}(l)\) at 1 bar near 0 °C, which signs are correct?

(A) ΔH < 0, ΔS < 0
(B) ΔH > 0, ΔS > 0
(C) ΔH > 0, ΔS < 0
(D) ΔH < 0, ΔS > 0

View Answer

Correct Answer: (B) Explanation: Melting absorbs heat and increases disorder.

44. Water has \(ΔH_{\text{vap}}=40.7\ \text{kJ·mol}^{-1}\) at its normal boiling point \(T_b=373\ \text{K}\). What is \(ΔS_{\text{vap}}\) at \(T_b\)?

(A) 40.7 J·mol\(^{-1}\)·K\(^{-1}\)
(B) 73.1 J·mol\(^{-1}\)·K\(^{-1}\)
(C) 109 J·mol\(^{-1}\)·K\(^{-1}\)
(D) 373 J·mol\(^{-1}\)·K\(^{-1}\)

View Answer

Correct Answer: (C) Explanation: At phase equilibrium, \(ΔG=0\Rightarrow ΔS=ΔH/T\). \(40{,}700/373\approx 109\ \text{J·mol}^{-1}\text{·K}^{-1}\).

45. If \(E^\circ = 0.000\ \text{V}\) for a redox reaction at 298 K, which statement is TRUE?

(A) \(ΔG^\circ<0\) and \(K>1\)
(B) \(ΔG^\circ=0\) and \(K=1\)
(C) \(ΔG^\circ>0\) and \(K<1\)
(D) Signs cannot be inferred

View Answer

Correct Answer: (B) Explanation: \(ΔG^\circ=−nFE^\circ=0\Rightarrow \ln K=−ΔG^\circ/(RT)=0\Rightarrow K=1\).

46. Using approximate standard molar entropies (J·mol⁻¹·K⁻¹): S°[N₂(g)]=192, S°[H₂(g)]=131, S°[NH₃(g)]=193. For
N₂(g) + 3 H₂(g) → 2 NH₃(g), what is ΔS°?

(A) −99 J·mol⁻¹·K⁻¹
(B) −199 J·mol⁻¹·K⁻¹
(C) +99 J·mol⁻¹·K⁻¹
(D) +199 J·mol⁻¹·K⁻¹

View Answer

Correct Answer: (B) Explanation: ΔS° = ΣS°(prod) − ΣS°(react) = 2(193) − [192 + 3(131)] = 386 − 585 = −199 J·mol⁻¹·K⁻¹.

47. A reaction has ΔH = −25.0 kJ·mol⁻¹ and ΔS = −50.0 J·mol⁻¹·K⁻¹. It is spontaneous at:

(A) T < 500 K
(B) T > 500 K
(C) Only at T = 500 K
(D) No temperatures

View Answer

Correct Answer: (A) Explanation: Threshold T* = ΔH/ΔS = (−25,000)/(−50) = 500 K. For T < T*, ΔG = ΔH − TΔS < 0.

48. At 298 K, ΔG° = +12.0 kJ·mol⁻¹. What is K? (R = 8.314 J·mol⁻¹·K⁻¹)

(A) 0.026
(B) 0.016
(C) 7.9×10⁻³
(D) 7.9×10⁻²

View Answer

Correct Answer: (C) Explanation: lnK = −ΔG°/(RT) = −12,000/(8.314×298) ≈ −4.84 ⇒ K ≈ e^−4.84 ≈ 7.9×10⁻³.

49. Compute ΔG° for CH₄(g)+2 O₂(g) → CO₂(g)+2 H₂O(l) using ΔGf° (kJ·mol⁻¹): CH₄=−50.8, CO₂=−394.4, H₂O(l)=−237.1, O₂=0.

(A) −742 kJ·mol⁻¹
(B) −818 kJ·mol⁻¹
(C) −890 kJ·mol⁻¹
(D) −938 kJ·mol⁻¹

View Answer

Correct Answer: (B) Explanation: Σprod − Σreact = [−394.4 + 2(−237.1)] − [−50.8] = −868.6 + 50.8 = −817.8 ≈ −818.

50. For NaCl(s) → Na⁺(aq) + Cl⁻(aq) at room temperature, the sign of ΔS° is most likely:

(A) Positive
(B) Negative
(C) Zero
(D) Cannot be defined at standard state

View Answer

Correct Answer: (A) Explanation: Dissolution increases dispersal of particles, typically giving ΔS° > 0.

51. A bomb calorimeter has C_cal = 3.15 kJ·K⁻¹ and contains 500 g water (c=4.184 J·g⁻¹·K⁻¹). Combustion of 1.00 g benzoic acid (M = 122.12 g·mol⁻¹) raises T by 2.00 K. Find ΔU_comb (kJ·mol⁻¹).

(A) −1.05×10³
(B) −1.18×10³
(C) −1.28×10³
(D) −1.38×10³

View Answer

Correct Answer: (C) Explanation: q_cal=3.15×2=6.30 kJ; q_water=500×4.184×2=4184 J=4.184 kJ; total = 10.484 kJ absorbed ⇒ q_rxn=−10.484 kJ for 1.00 g. n = 1.00/122.12 = 8.19×10⁻³ mol ⇒ ΔU ≈ −10.484/8.19×10⁻³ ≈ −1.28×10³ kJ·mol⁻¹.

52. Estimate ΔH for N₂ + 3 H₂ → 2 NH₃ using average bond energies (kJ·mol⁻¹): D(N≡N)=945, D(H–H)=436, D(N–H)=391.

(A) −23
(B) −93
(C) +93
(D) +323

View Answer

Correct Answer: (B) Explanation: Broken = 945 + 3×436 = 2253; formed = 2×3×391 = 2346; ΔH ≈ 2253 − 2346 = −93 kJ·mol⁻¹.

53. For 1 mol of an ideal gas at the same temperature, the molar entropy is larger at:

(A) 2 atm
(B) 1 atm
(C) Equal at both pressures
(D) Depends on gas identity only

View Answer

Correct Answer: (B) Explanation: S increases with volume (decreases with pressure) at constant T for an ideal gas.

54. For a galvanic reaction with n = 2 and E° = 1.10 V at 298 K, the maximum electrical work for 0.50 mol of reaction advancement is approximately:

(A) −53.0 kJ
(B) −106 kJ
(C) −212 kJ
(D) +106 kJ

View Answer

Correct Answer: (B) Explanation: w_max=−n F E° × (extent) = −(2)(96485)(1.10)×0.50 ≈ −1.06×10⁵ J = −106 kJ.

55. Which statement about spontaneity is TRUE?

(A) If ΔG° > 0, the reaction cannot be spontaneous under any conditions.
(B) If ΔG° < 0, ΔG is always negative for any Q.
(C) A reaction with ΔG° > 0 can be spontaneous if Q is sufficiently small so that ΔG = ΔG° + RT lnQ < 0.
(D) ΔG does not depend on Q.

View Answer

Correct Answer: (C) Explanation: Nonstandard free energy: ΔG = ΔG° + RT lnQ. Small Q (reactant-heavy) can make ΔG negative even if ΔG° > 0.

Ap Chemistry 2차 - Unit 6

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