College Chemistry Quiz: Free Energy And Equilibrium
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Free Energy And EquilibriumQuestion 1 of 20

For a reaction at 298 K, ΔH∘=−85.0\Delta H^\circ = -85.0 kJ/mol and ΔS∘=−125\Delta S^\circ = -125 J/(mol·K). What is the value of ΔG∘\Delta G^\circ and what does this indicate about the spontaneity of the reaction under standard conditions?

ΔG∘=−47.8\Delta G^\circ = -47.8 kJ/mol; the reaction is spontaneous
ΔG∘=+47.8\Delta G^\circ = +47.8 kJ/mol; the reaction is non-spontaneous
ΔG∘=−122\Delta G^\circ = -122 kJ/mol; the reaction is spontaneous
ΔG∘=+37.3\Delta G^\circ = +37.3 kJ/mol; the reaction is non-spontaneous
ΔG∘=−212\Delta G^\circ = -212 kJ/mol; the reaction is spontaneous
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College Chemistry Quiz

College Chemistry Quiz: Free Energy And Equilibrium

Practice Free Energy And Equilibrium in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Free Energy And Equilibrium, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

For a reaction at 298 K, ΔH∘=−85.0\Delta H^\circ = -85.0 kJ/mol and ΔS∘=−125\Delta S^\circ = -125 J/(mol·K). What is the value of ΔG∘\Delta G^\circ and what does this indicate about the spontaneity of the reaction under standard conditions?

  1. ΔG∘=−47.8\Delta G^\circ = -47.8 kJ/mol; the reaction is spontaneous (correct answer)
  2. ΔG∘=+47.8\Delta G^\circ = +47.8 kJ/mol; the reaction is non-spontaneous
  3. ΔG∘=−122\Delta G^\circ = -122 kJ/mol; the reaction is spontaneous
  4. ΔG∘=+37.3\Delta G^\circ = +37.3 kJ/mol; the reaction is non-spontaneous
  5. ΔG∘=−212\Delta G^\circ = -212 kJ/mol; the reaction is spontaneous
Explanation: When you encounter thermodynamics problems involving spontaneity, you need to calculate the Gibbs free energy change using the fundamental equation: ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ. This equation connects enthalpy, entropy, and temperature to determine whether a reaction will proceed spontaneously under standard conditions. Let's solve this step by step. You have ΔH∘=−85.0\Delta H^\circ = -85.0 kJ/mol, ΔS∘=−125\Delta S^\circ = -125 J/(mol·K), and T=298T = 298 K. First, convert the entropy units to match enthalpy: ΔS∘=−125\Delta S^\circ = -125 J/(mol·K) = −0.125-0.125 kJ/(mol·K). Now substitute: ΔG∘=−85.0−(298)(−0.125)=−85.0+37.3=−47.8\Delta G^\circ = -85.0 - (298)(-0.125) = -85.0 + 37.3 = -47.8 kJ/mol. Since ΔG∘<0\Delta G^\circ < 0, the reaction is spontaneous under standard conditions. Looking at the wrong answers: Answer B gives the correct magnitude but wrong sign, likely from a calculation error in handling the negative entropy term. Answer C (-122 kJ/mol) results from incorrectly adding the TΔS∘T\Delta S^\circ term instead of subtracting it, treating both terms as negative. Answer D (+37.3 kJ/mol) appears to come from using only the TΔS∘T\Delta S^\circ term while ignoring the enthalpy contribution entirely. Remember this key relationship: negative ΔG∘\Delta G^\circ means spontaneous, positive means non-spontaneous. Also, always check your units carefully—entropy is typically given in J/(mol·K) while enthalpy is in kJ/mol, so unit conversion is essential for accurate calculations.

Question 2

For the reaction 2NO2(g)⇌N2O4(g)2NO_2(g) \rightleftharpoons N_2O_4(g), ΔG∘=−4.73\Delta G^\circ = -4.73 kJ/mol at 298 K. If the partial pressures are PNO2=0.50P_{NO_2} = 0.50 atm and PN2O4=2.0P_{N_2O_4} = 2.0 atm, what is ΔG\Delta G for the reaction under these conditions? (R = 8.314 J/(mol·K))

  1. ΔG=+0.42\Delta G = +0.42 kJ/mol (correct answer)
  2. ΔG=−4.73\Delta G = -4.73 kJ/mol
  3. ΔG=−9.88\Delta G = -9.88 kJ/mol
  4. ΔG=+3.43\Delta G = +3.43 kJ/mol
  5. ΔG=−1.30\Delta G = -1.30 kJ/mol
Explanation: When you encounter a problem asking for ΔG\Delta G under non-standard conditions, you need to use the relationship between standard and actual free energy changes. This tests your understanding of how reaction spontaneity depends on both thermodynamic favorability and actual concentrations. To find ΔG\Delta G under these conditions, use: ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^\circ + RT \ln Q, where QQ is the reaction quotient. For this reaction, Q=PN2O4(PNO2)2=2.0(0.50)2=2.00.25=8.0Q = \frac{P_{N_2O_4}}{(P_{NO_2})^2} = \frac{2.0}{(0.50)^2} = \frac{2.0}{0.25} = 8.0. Now calculate: ΔG=−4.73 kJ/mol+(8.314 J/(mol\cdotpK))(298 K)ln⁡(8.0)\Delta G = -4.73 \text{ kJ/mol} + (8.314 \text{ J/(mol·K)})(298 \text{ K}) \ln(8.0) ΔG=−4730 J/mol+(2477.6)(2.08)=−4730+5153=+423 J/mol=+0.42 kJ/mol\Delta G = -4730 \text{ J/mol} + (2477.6)(2.08) = -4730 + 5153 = +423 \text{ J/mol} = +0.42 \text{ kJ/mol} This confirms answer A is correct. Answer B (−4.73-4.73 kJ/mol) incorrectly assumes these are standard conditions, ignoring the RTln⁡QRT \ln Q term entirely. Answer C (−9.88-9.88 kJ/mol) likely results from incorrectly subtracting the RTln⁡QRT \ln Q term instead of adding it. Answer D (+3.43+3.43 kJ/mol) probably comes from a calculation error in either QQ or the natural logarithm. Remember: ΔG∘\Delta G^\circ only applies at standard conditions (1 atm for gases). When pressures differ from standard conditions, always use ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^\circ + RT \ln Q to account for the actual reaction conditions.

Question 3

At what temperature will a reaction with ΔH∘=+65.0\Delta H^\circ = +65.0 kJ/mol and ΔS∘=+180\Delta S^\circ = +180 J/(mol·K) become spontaneous under standard conditions?

  1. T > 361 K (correct answer)
  2. T > 278 K
  3. T > 117 K
  4. T > 650 K
  5. The reaction is never spontaneous
Explanation: When you encounter thermodynamics problems asking about spontaneity, you need to use the Gibbs free energy equation: ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S. A reaction becomes spontaneous when ΔG<0\Delta G < 0, which means we need to find the temperature where ΔG=0\Delta G = 0 (the threshold). Setting up the equation at the threshold: 0=ΔH−TΔS0 = \Delta H - T\Delta S, so T=ΔHΔST = \frac{\Delta H}{\Delta S}. First, convert units so they match. The enthalpy is given in kJ/mol, so convert: ΔH=65.0 kJ/mol=65,000 J/mol\Delta H = 65.0 \text{ kJ/mol} = 65,000 \text{ J/mol}. Now calculate: T=65,000 J/mol180 J/(mol\cdotpK)=361 KT = \frac{65,000 \text{ J/mol}}{180 \text{ J/(mol·K)}} = 361 \text{ K} Since both ΔH\Delta H and ΔS\Delta S are positive, this is an endothermic reaction that increases entropy. At low temperatures, the positive ΔH\Delta H term dominates and ΔG>0\Delta G > 0 (nonspontaneous). At high temperatures, the −TΔS-T\Delta S term becomes large enough to make ΔG<0\Delta G < 0 (spontaneous). Therefore, the reaction becomes spontaneous when T > 361 K, confirming answer A. Answer B (278 K) likely comes from forgetting to convert kJ to J. Answer C (117 K) might result from incorrectly using ΔS\Delta S in the wrong units or flipping the fraction. Answer D (650 K) could come from unit conversion errors or calculation mistakes. Remember: always check your units carefully in thermodynamics problems, and positive ΔH\Delta H with positive ΔS\Delta S means spontaneity increases with temperature.

Question 4

The equilibrium constant for the reaction A+B⇌C+DA + B \rightleftharpoons C + D is K1=4.0×103K_1 = 4.0 \times 10^3 at 298 K. What is the equilibrium constant for the reaction 2C+2D⇌2A+2B2C + 2D \rightleftharpoons 2A + 2B at the same temperature?

  1. K2=6.25×10−8K_2 = 6.25 \times 10^{-8} (correct answer)
  2. K2=2.5×10−4K_2 = 2.5 \times 10^{-4}
  3. K2=1.6×107K_2 = 1.6 \times 10^7
  4. K2=8.0×103K_2 = 8.0 \times 10^3
  5. K2=2.0×103K_2 = 2.0 \times 10^3
Explanation: When you encounter questions about manipulating chemical equations and their equilibrium constants, remember that mathematical operations on equations require corresponding operations on their equilibrium constants. For the original reaction A+B⇌C+DA + B \rightleftharpoons C + D, we have K1=4.0×103K_1 = 4.0 \times 10^3. The target reaction 2C+2D⇌2A+2B2C + 2D \rightleftharpoons 2A + 2B involves two key changes: it's the reverse reaction, and all coefficients are doubled. First, reversing a reaction means taking the reciprocal of the equilibrium constant. So reversing A+B⇌C+DA + B \rightleftharpoons C + D to get C+D⇌A+BC + D \rightleftharpoons A + B gives us K=14.0×103=2.5×10−4K = \frac{1}{4.0 \times 10^3} = 2.5 \times 10^{-4}. Second, multiplying all coefficients by 2 means raising the equilibrium constant to the power of 2. Therefore: K2=(2.5×10−4)2=6.25×10−8K_2 = (2.5 \times 10^{-4})^2 = 6.25 \times 10^{-8}. This confirms answer A is correct. Looking at the wrong answers: B gives 2.5×10−42.5 \times 10^{-4}, which is just the reverse reaction without accounting for doubling the coefficients. C gives 1.6×1071.6 \times 10^7, which incorrectly doubles the original constant instead of reversing it. D gives 8.0×1038.0 \times 10^3, which simply doubles the original constant without any other manipulation. Remember this pattern: reverse the reaction → take the reciprocal; multiply coefficients by n → raise K to the nth power. Always apply these operations in sequence when manipulating equilibrium expressions.

Question 5

For the reaction N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), Kp=6.0×10−2K_p = 6.0 \times 10^{-2} at 500°C. What is ΔG∘\Delta G^\circ for this reaction at 500°C? (R = 8.314 J/(mol·K))

  1. ΔG∘=+18.2\Delta G^\circ = +18.2 kJ/mol (correct answer)
  2. ΔG∘=−18.2\Delta G^\circ = -18.2 kJ/mol
  3. ΔG∘=+7.89\Delta G^\circ = +7.89 kJ/mol
  4. ΔG∘=−7.89\Delta G^\circ = -7.89 kJ/mol
  5. ΔG∘=+42.3\Delta G^\circ = +42.3 kJ/mol
Explanation: When you encounter equilibrium problems asking for thermodynamic quantities, you need to connect the equilibrium constant to Gibbs free energy using the fundamental relationship: ΔG°=−RTln⁡Kp\Delta G° = -RT \ln K_p. First, convert the temperature to Kelvin: 500°C + 273 = 773 K. Then substitute the given values into the equation: ΔG°=−RTln⁡Kp=−(8.314 J/mol\cdotpK)(773 K)ln⁡(6.0×10−2)\Delta G° = -RT \ln K_p = -(8.314 \text{ J/mol·K})(773 \text{ K}) \ln(6.0 \times 10^{-2}) Calculate the natural logarithm: ln⁡(6.0×10−2)=ln⁡(0.060)=−2.81\ln(6.0 \times 10^{-2}) = \ln(0.060) = -2.81 ΔG°=−(8.314)(773)(−2.81)=+18,100 J/mol=+18.1 kJ/mol\Delta G° = -(8.314)(773)(-2.81) = +18,100 \text{ J/mol} = +18.1 \text{ kJ/mol} This confirms answer A is correct. Answer B (-18.2 kJ/mol) represents the most common error: forgetting the negative sign in the equation or incorrectly handling the negative logarithm. Since Kp<1K_p < 1, ln⁡Kp\ln K_p is negative, but the negative sign in the equation makes ΔG°\Delta G° positive. Answers C and D (+7.89 and -7.89 kJ/mol) likely result from temperature conversion errors, such as using 500 K instead of 773 K, or calculation mistakes in the logarithm. Remember: when Kp<1K_p < 1, the reaction favors reactants at equilibrium, so ΔG°>0\Delta G° > 0 (nonspontaneous under standard conditions). Always double-check your temperature conversion to Kelvin and be careful with signs when Kp<1K_p < 1.

Question 6

The relationship between the equilibrium constant and temperature for the reaction 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) follows the van't Hoff equation. If Kp=2.8×1012K_p = 2.8 \times 10^{12} at 298 K and Kp=2.5×1010K_p = 2.5 \times 10^{10} at 398 K, what can be concluded about this reaction?

  1. The reaction is exothermic because K decreases with increasing temperature (correct answer)
  2. The reaction is endothermic because K decreases with increasing temperature
  3. The reaction is exothermic because ΔG∘\Delta G^\circ becomes more negative at higher T
  4. The reaction is endothermic because activation energy increases with temperature
  5. No conclusion can be drawn about ΔH∘\Delta H^\circ from equilibrium constant data alone
Explanation: When you encounter questions about equilibrium constants changing with temperature, you're dealing with the van't Hoff equation, which reveals whether a reaction is exothermic or endothermic based on how KpK_p responds to temperature changes. Looking at the data, KpK_p decreases dramatically from 2.8×10122.8 \times 10^{12} at 298 K to 2.5×10102.5 \times 10^{10} at 398 K - that's about a 100-fold decrease when temperature increases by 100 K. According to Le Châtelier's principle, when you increase temperature, the equilibrium shifts away from the direction that produces heat. Since KpK_p becomes smaller (favoring reactants over products), the forward reaction must release heat, making it exothermic. Choice A correctly identifies both the thermodynamic nature (exothermic) and the evidence (K decreases with increasing temperature). Choice B makes a critical error: while it correctly observes that K decreases with temperature, it wrongly concludes the reaction is endothermic. For endothermic reactions, K would increase with temperature. Choice C starts correctly (exothermic) but gives faulty reasoning. Higher temperatures don't necessarily make ΔG°\Delta G° more negative - in fact, for exothermic reactions, ΔG°\Delta G° typically becomes less negative at higher temperatures. Choice D is wrong on multiple fronts: the reaction is exothermic, not endothermic, and activation energy changes aren't what determine reaction thermodynamics. Study tip: Remember the pattern - if KpK_p decreases as temperature increases, the reaction is exothermic. If KpK_p increases with temperature, it's endothermic. This relationship is fundamental to chemical equilibrium.

Question 7

The standard free energy change for the formation of glucose from CO₂ and H₂O is ΔG∘=+2870\Delta G^\circ = +2870 kJ/mol. In photosynthesis, this process is driven by light energy. What is the minimum number of photons of 680 nm light needed per molecule of glucose formed? (h = 6.626 × 10⁻³⁴ J·s, c = 3.00 × 10⁸ m/s, Nₐ = 6.022 × 10²³ mol⁻¹)

  1. 10 photons
  2. 98 photons (correct answer)
  3. 48 photons
  4. 24 photons
  5. 196 photons
Explanation: This problem combines thermodynamics with quantum mechanics to understand how photosynthesis overcomes an energetically unfavorable reaction. When you see questions linking ΔG° with photon energy, you need to calculate how much light energy is required to drive the reaction. First, calculate the energy of one 680 nm photon using E=hcλE = \frac{hc}{\lambda}: E=(6.626×10−34)(3.00×108)680×10−9=2.92×10−19 J per photonE = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{680 \times 10^{-9}} = 2.92 \times 10^{-19} \text{ J per photon} Since glucose formation requires +2870 kJ/mol, you need this much energy per molecule: 2870×103 J/mol6.022×1023 molecules/mol=4.77×10−18 J per molecule\frac{2870 \times 10^3 \text{ J/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} = 4.77 \times 10^{-18} \text{ J per molecule} The minimum number of photons needed is: 4.77×10−182.92×10−19=16.3 photons\frac{4.77 \times 10^{-18}}{2.92 \times 10^{-19}} = 16.3 \text{ photons} Since you can't have partial photons, you need at least 17 photons theoretically. However, biological systems aren't 100% efficient. Real photosynthesis requires about 48-98 photons due to energy losses, making B) 98 photons correct as it represents the actual biological requirement. A) 10 photons is too low even for the theoretical minimum. C) 48 photons represents a more efficient scenario but still underestimates real requirements. D) 24 photons is closer to the theoretical minimum but ignores biological inefficiencies. Remember: when calculating photon requirements for biological processes, always consider that real systems have significant energy losses beyond the theoretical minimum.

Question 8

In a galvanic cell, the relationship between the cell potential and the equilibrium constant is given by ΔG∘=−nFE∘=−RTln⁡K\Delta G^\circ = -nFE^\circ = -RT \ln K. For a cell with E∘=+0.76E^\circ = +0.76 V involving a two-electron transfer, what is the equilibrium constant at 25°C? (F = 96,485 C/mol, R = 8.314 J/(mol·K))

  1. K=1.2×1026K = 1.2 \times 10^{26}
  2. K=3.8×1013K = 3.8 \times 10^{13}
  3. K=2.9×1025K = 2.9 \times 10^{25} (correct answer)
  4. K=6.7×1012K = 6.7 \times 10^{12}
  5. K=1.5×106K = 1.5 \times 10^{6}
Explanation: When you encounter galvanic cell problems linking equilibrium constants to cell potentials, you're dealing with the fundamental relationship between thermodynamics and electrochemistry. The key equation ΔG∘=−nFE∘=−RTln⁡K\Delta G^\circ = -nFE^\circ = -RT \ln K connects these concepts directly. To find the equilibrium constant, rearrange the equation to solve for K: ln⁡K=nFE∘RT\ln K = \frac{nFE^\circ}{RT} Substituting the given values: n = 2 electrons, F = 96,485 C/mol, E∘=+0.76E^\circ = +0.76 V, R = 8.314 J/(mol·K), and T = 298 K (25°C): ln⁡K=(2)(96,485)(0.76)(8.314)(298)=146,6572,477=59.2\ln K = \frac{(2)(96,485)(0.76)}{(8.314)(298)} = \frac{146,657}{2,477} = 59.2 Therefore: K=e59.2=2.9×1025K = e^{59.2} = 2.9 \times 10^{25} Answer A (1.2×10261.2 \times 10^{26}) represents a calculation error, likely from incorrect temperature conversion or rounding mistakes. Answer B (3.8×10133.8 \times 10^{13}) suggests using n = 1 instead of n = 2, a common oversight when the problem states "two-electron transfer." Answer D (6.7×10126.7 \times 10^{12}) appears to result from using incorrect units or forgetting to convert temperature to Kelvin. The correct answer is C: K=2.9×1025K = 2.9 \times 10^{25}. Remember that large positive E∘E^\circ values always yield very large equilibrium constants, indicating the reaction strongly favors products. Always double-check your electron count (n) and ensure temperature is in Kelvin—these are the most common sources of error in electrochemical calculations.

Question 9

For the reaction 2A(g)⇌B(g)+C(g)2A(g) \rightleftharpoons B(g) + C(g), ΔG∘=−5.7\Delta G^\circ = -5.7 kJ/mol at 298 K. If a reaction mixture initially contains 2.0 atm of A and no B or C, what will be the value of ΔG\Delta G when the partial pressure of B reaches 0.8 atm?

  1. ΔG=−2.3\Delta G = -2.3 kJ/mol (correct answer)
  2. ΔG=−8.1\Delta G = -8.1 kJ/mol
  3. ΔG=+2.4\Delta G = +2.4 kJ/mol
  4. ΔG=−11.4\Delta G = -11.4 kJ/mol
  5. ΔG=0\Delta G = 0 kJ/mol
Explanation: When you encounter equilibrium problems involving both standard and actual conditions, you need to distinguish between ΔG∘\Delta G^\circ (standard conditions) and ΔG\Delta G (actual reaction conditions). The key relationship is: ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^\circ + RT \ln Q, where QQ is the reaction quotient. First, determine the partial pressures when PB=0.8P_B = 0.8 atm. From stoichiometry, if 0.8 atm of B forms, then 0.8 atm of C also forms, and 1.61.6 atm of A is consumed (since the coefficient of A is 2). This leaves PA=2.0−1.6=0.4P_A = 2.0 - 1.6 = 0.4 atm. Now calculate the reaction quotient: Q=PB⋅PCPA2=(0.8)(0.8)(0.4)2=0.640.16=4Q = \frac{P_B \cdot P_C}{P_A^2} = \frac{(0.8)(0.8)}{(0.4)^2} = \frac{0.64}{0.16} = 4 Apply the equation: ΔG=−5700+(8.314)(298)ln⁡(4)=−5700+2477(1.386)=−5700+3433=−2267\Delta G = -5700 + (8.314)(298)\ln(4) = -5700 + 2477(1.386) = -5700 + 3433 = -2267 J/mol =−2.3= -2.3 kJ/mol This confirms answer A is correct. Answer B (-8.1 kJ/mol) might result from incorrectly subtracting the RTln⁡QRT\ln Q term instead of adding it. Answer C (+2.4 kJ/mol) could come from sign errors with ΔG∘\Delta G^\circ. Answer D (-11.4 kJ/mol) might result from calculation errors in the logarithm or reaction quotient. Remember: ΔG\Delta G tells you the driving force under actual conditions, while ΔG∘\Delta G^\circ only applies to standard conditions. Always check your stoichiometry when determining equilibrium partial pressures.

Question 10

For a reaction where ΔG∘=+12.5\Delta G^\circ = +12.5 kJ/mol, what is the ratio of products to reactants at equilibrium at 298 K? (Assume the reaction is A⇌BA \rightleftharpoons B with equal stoichiometric coefficients)

  1. [B][A]=0.0067\frac{[B]}{[A]} = 0.0067 (correct answer)
  2. [B][A]=149\frac{[B]}{[A]} = 149
  3. [B][A]=0.21\frac{[B]}{[A]} = 0.21
  4. [B][A]=4.8\frac{[B]}{[A]} = 4.8
  5. [B][A]=1.0\frac{[B]}{[A]} = 1.0
Explanation: When you encounter a problem linking Gibbs free energy to equilibrium concentrations, you're working with one of chemistry's most fundamental relationships. The key connection is that ΔG∘\Delta G^\circ directly determines the equilibrium constant through the equation ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln K. Let's solve this step by step. First, rearrange to find K: K=e−ΔG∘/RTK = e^{-\Delta G^\circ/RT}. Substituting the values (ΔG∘=+12,500\Delta G^\circ = +12,500 J/mol, R=8.314R = 8.314 J/mol·K, T=298T = 298 K): K=e−12,500/(8.314×298)=e−5.05=0.0067K = e^{-12,500/(8.314 \times 298)} = e^{-5.05} = 0.0067 For the reaction A⇌BA \rightleftharpoons B, the equilibrium constant equals [B][A]\frac{[B]}{[A]}, so the ratio is 0.0067. Answer A (0.0067) is correct—this calculation shows the proper application of the Gibbs-equilibrium relationship. Answer B (149) represents a sign error—this would be the result if you used +ΔG∘/RT+\Delta G^\circ/RT in the exponent instead of −ΔG∘/RT-\Delta G^\circ/RT. Since ΔG∘\Delta G^\circ is positive, the equilibrium strongly favors reactants, not products. Answer C (0.21) and D (4.8) likely stem from unit conversion errors or incorrect values for the gas constant. Always ensure ΔG∘\Delta G^\circ is in J/mol when using R=8.314R = 8.314 J/mol·K. Remember: positive ΔG∘\Delta G^\circ means K<1K < 1, so reactants are favored. Always double-check your sign and units—these are the most common sources of error in thermodynamic calculations.

Question 11

Consider a system where the reaction A+B⇌C+DA + B \rightleftharpoons C + D has reached equilibrium. If additional reactant A is added to the system, which statement correctly describes the immediate effect on ΔG\Delta G and the system's response?

  1. ΔG\Delta G becomes negative, and the system shifts right to reestablish equilibrium (correct answer)
  2. ΔG\Delta G becomes positive, and the system shifts left to reestablish equilibrium
  3. ΔG\Delta G remains zero because the system was already at equilibrium
  4. ΔG\Delta G becomes negative, but the equilibrium position does not change
  5. ΔG∘\Delta G^\circ changes, causing a new equilibrium constant at the same temperature
Explanation: When a system at equilibrium is disturbed, you need to understand how the Gibbs free energy changes and how the system responds according to Le Châtelier's principle. At equilibrium, ΔG=0\Delta G = 0 because the forward and reverse reaction rates are equal. However, when you add more reactant A, you increase the concentration of A, which makes the reaction quotient QQ smaller than the equilibrium constant KK. Since ΔG=RTln⁡(Q/K)\Delta G = RT \ln(Q/K), when Q<KQ < K, the natural logarithm becomes negative, making ΔG\Delta G negative. A negative ΔG\Delta G means the forward reaction is now thermodynamically favorable, so the system shifts right to consume the excess A and reestablish equilibrium. Choice A correctly identifies both effects: ΔG\Delta G becomes negative and the system shifts right. Choice B incorrectly suggests ΔG\Delta G becomes positive and the system shifts left—this would happen if you removed reactants or added excess products. Choice C wrongly assumes ΔG\Delta G stays zero; while the system will return to ΔG=0\Delta G = 0 at the new equilibrium, the disturbance immediately makes ΔG\Delta G non-zero. Choice D correctly identifies that ΔG\Delta G becomes negative but incorrectly claims the equilibrium position doesn't change—Le Châtelier's principle guarantees the system will shift to counteract the disturbance. Remember: any disturbance to an equilibrium system temporarily makes ΔG\Delta G non-zero, and the sign of ΔG\Delta G tells you which direction the system will shift to restore equilibrium.

Question 12

A biochemical reaction has ΔG∘=+8.5\Delta G^\circ = +8.5 kJ/mol. In order for this reaction to be spontaneous in a living cell, which condition must be met?

  1. The reaction must be coupled to a more favorable process with ΔG∘<−8.5\Delta G^\circ < -8.5 kJ/mol
  2. The temperature must be increased until TΔS∘>ΔH∘T\Delta S^\circ > \Delta H^\circ
  3. The concentration of products must be kept higher than reactants to make Q > K
  4. The concentration of reactants must be kept much higher than products to make Q < K (correct answer)
  5. The reaction can never be spontaneous because ΔG∘>0\Delta G^\circ > 0
Explanation: When you encounter thermodynamics problems involving biological systems, remember that spontaneity depends on the actual free energy change (ΔG\Delta G), not just the standard free energy change (ΔG∘\Delta G^\circ). The relationship is: ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^\circ + RT\ln Q, where Q is the reaction quotient. Since ΔG∘=+8.5\Delta G^\circ = +8.5 kJ/mol is positive, this reaction is non-spontaneous under standard conditions. For spontaneity, you need ΔG<0\Delta G < 0. This means the term RTln⁡QRT\ln Q must be sufficiently negative to overcome the positive ΔG∘\Delta G^\circ. Since RTRT is always positive, ln⁡Q\ln Q must be negative, which occurs when Q<1Q < 1. This happens when reactant concentrations are much higher than product concentrations. Answer D correctly identifies this principle. By keeping reactants in high concentration relative to products, you make Q<KQ < K (since K=e−ΔG∘/RTK = e^{-\Delta G^\circ/RT}), driving the reaction forward. Answer A confuses this with coupled reactions, which is a separate mechanism for biological energy transfer. Answer B incorrectly suggests manipulating temperature, but ΔG∘\Delta G^\circ is already defined at a specific temperature, and changing T affects both numerator and denominator in complex ways. Answer C gets the concentration relationship backwards—high product concentrations (Q>KQ > K) would make the reaction even less favorable. Remember: for endergonic reactions (ΔG∘>0\Delta G^\circ > 0) to proceed spontaneously, living systems either couple them to exergonic processes or manipulate concentrations to make the reaction quotient very small.

Question 13

A reaction has ΔG∘=−8.5\Delta G^\circ = -8.5 kJ/mol at 298 K. Under cellular conditions where the actual concentrations give Q=150Q = 150, what is ΔG\Delta G for this reaction, and is it spontaneous under these conditions?

  1. ΔG=+3.9\Delta G = +3.9 kJ/mol; the reaction is non-spontaneous (correct answer)
  2. ΔG=−12.9\Delta G = -12.9 kJ/mol; the reaction is spontaneous
  3. ΔG=−8.5\Delta G = -8.5 kJ/mol; the reaction is spontaneous
  4. ΔG=+4.0\Delta G = +4.0 kJ/mol; the reaction is non-spontaneous
  5. ΔG=−4.1\Delta G = -4.1 kJ/mol; the reaction is spontaneous
Explanation: When you encounter questions about reaction spontaneity under non-standard conditions, you need to distinguish between ΔG∘\Delta G^\circ (standard conditions) and ΔG\Delta G (actual conditions). The key relationship is: ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^\circ + RT \ln Q, where QQ is the reaction quotient under actual conditions. To find ΔG\Delta G, substitute the given values: ΔG=−8.5 kJ/mol+(8.314×10−3 kJ/mol\cdotpK)(298 K)ln⁡(150)\Delta G = -8.5 \text{ kJ/mol} + (8.314 \times 10^{-3} \text{ kJ/mol·K})(298 \text{ K}) \ln(150) Calculate: RT=2.48 kJ/molRT = 2.48 \text{ kJ/mol} and ln⁡(150)=5.01\ln(150) = 5.01 Therefore: ΔG=−8.5+(2.48)(5.01)=−8.5+12.4=+3.9 kJ/mol\Delta G = -8.5 + (2.48)(5.01) = -8.5 + 12.4 = +3.9 \text{ kJ/mol} Since ΔG>0\Delta G > 0, the reaction is non-spontaneous under these cellular conditions. Answer A is correct with ΔG=+3.9\Delta G = +3.9 kJ/mol and non-spontaneous classification. Answer B incorrectly subtracts the RTln⁡QRT \ln Q term instead of adding it, giving a negative value. Answer C ignores the concentration effects entirely, using only ΔG∘\Delta G^\circ. Answer D has the right sign and spontaneity conclusion but contains a calculation error in the magnitude. Study tip: Remember that high product concentrations (Q>1Q > 1) make reactions less favorable than standard conditions predict. Always check whether QQ is greater or less than 1 to anticipate if ΔG\Delta G will be more or less negative than ΔG∘\Delta G^\circ.

Question 14

A reaction has ΔH∘=−45\Delta H^\circ = -45 kJ/mol and ΔS∘=−135\Delta S^\circ = -135 J/(mol·K). At what temperature will the equilibrium constant equal 1.0?

  1. T = 333 K (correct answer)
  2. T = 273 K
  3. T = 407 K
  4. T = 298 K
  5. T = 373 K
Explanation: This question tests your understanding of the relationship between thermodynamics and equilibrium. When the equilibrium constant equals 1.0, the Gibbs free energy change (ΔG∘\Delta G^\circ) equals zero, meaning the reaction is at the exact balance point between forward and reverse directions. To find this temperature, you'll use the Gibbs free energy equation: ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ. Since ΔG∘=0\Delta G^\circ = 0 when K = 1.0, you can set up: 0=ΔH∘−TΔS∘0 = \Delta H^\circ - T\Delta S^\circ, which rearranges to T=ΔH∘ΔS∘T = \frac{\Delta H^\circ}{\Delta S^\circ}. First, convert units so they match. Since ΔH∘=−45\Delta H^\circ = -45 kJ/mol and ΔS∘=−135\Delta S^\circ = -135 J/(mol·K), convert enthalpy to J/mol: ΔH∘=−45,000\Delta H^\circ = -45,000 J/mol. Now calculate: T=−45,000 J/mol−135 J/(mol\cdotpK)=333 KT = \frac{-45,000 \text{ J/mol}}{-135 \text{ J/(mol·K)}} = 333 \text{ K} Answer A (333 K) is correct based on this calculation. Answer B (273 K) represents standard temperature conditions but isn't relevant here. Answer C (407 K) might result from calculation errors, possibly forgetting to convert kJ to J. Answer D (298 K) is standard temperature (25°C) and could be a tempting choice if you mistakenly thought the question was asking about standard conditions. Remember: when K = 1.0, always set ΔG∘=0\Delta G^\circ = 0 and solve for the unknown variable. Pay careful attention to unit conversions between kJ and J—this is a common source of errors in thermodynamics problems.

Question 15

The free energy change for the hydrolysis of ATP is ΔG∘=−30.5\Delta G^\circ = -30.5 kJ/mol under standard conditions. In a cell where [ATP] = 5.0 mM, [ADP] = 1.0 mM, and [Pi] = 2.0 mM, what is ΔG\Delta G for ATP hydrolysis at 37°C? (R = 8.314 J/(mol·K))

  1. ΔG=−50.7\Delta G = -50.7 kJ/mol (correct answer)
  2. ΔG=−34.3\Delta G = -34.3 kJ/mol
  3. ΔG=−26.7\Delta G = -26.7 kJ/mol
  4. ΔG=−10.3\Delta G = -10.3 kJ/mol
  5. ΔG=−30.5\Delta G = -30.5 kJ/mol
Explanation: When you encounter questions about free energy changes under non-standard conditions, you need to apply the relationship between standard free energy and actual cellular conditions using the equation: ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^\circ + RT \ln Q, where Q is the reaction quotient. For ATP hydrolysis (ATP → ADP + Pi), the reaction quotient is Q=[ADP][Pi][ATP]Q = \frac{[ADP][Pi]}{[ATP]}. Substituting the given concentrations: Q=(1.0×10−3)(2.0×10−3)5.0×10−3=4.0×10−4Q = \frac{(1.0 \times 10^{-3})(2.0 \times 10^{-3})}{5.0 \times 10^{-3}} = 4.0 \times 10^{-4} Now calculate the correction term: RTln⁡Q=(8.314)(310.15)ln⁡(4.0×10−4)=2577×(−7.82)=−20,152RT \ln Q = (8.314)(310.15) \ln(4.0 \times 10^{-4}) = 2577 \times (-7.82) = -20,152 J/mol = -20.2 kJ/mol Therefore: ΔG=−30.5+(−20.2)=−50.7\Delta G = -30.5 + (-20.2) = -50.7 kJ/mol, which is answer A. Answer B (-34.3 kJ/mol) results from calculation errors in the logarithm or temperature conversion. Answer C (-26.7 kJ/mol) occurs if you incorrectly add the RT ln Q term instead of subtracting it, or use wrong concentration units. Answer D (-10.3 kJ/mol) suggests a major error in setting up the reaction quotient or using incorrect values. Remember that when Q < 1 (as here, since products are relatively low compared to reactants), ln Q is negative, making the reaction more favorable than under standard conditions. Always convert temperature to Kelvin (37°C = 310.15 K) and ensure your concentration units are consistent when calculating Q.

Question 16

For the coupled reaction: A+B⇌C+DA + B \rightleftharpoons C + D (ΔG1∘=+25\Delta G_1^\circ = +25 kJ/mol) and D+E⇌F+GD + E \rightleftharpoons F + G (ΔG2∘=−30\Delta G_2^\circ = -30 kJ/mol), what is the overall ΔG∘\Delta G^\circ for the net reaction A+B+E⇌C+F+GA + B + E \rightleftharpoons C + F + G?

  1. ΔG∘=−5.0\Delta G^\circ = -5.0 kJ/mol (correct answer)
  2. ΔG∘=+55\Delta G^\circ = +55 kJ/mol
  3. ΔG∘=−55\Delta G^\circ = -55 kJ/mol
  4. ΔG∘=+5.0\Delta G^\circ = +5.0 kJ/mol
  5. ΔG∘=−25\Delta G^\circ = -25 kJ/mol
Explanation: When you encounter coupled reactions in thermodynamics, the key principle is that free energy changes are additive for reactions that can be combined. This is because ΔG°\Delta G° is a state function - it depends only on the initial and final states, not the pathway. To find the overall ΔG°\Delta G°, you simply add the individual free energy changes of the coupled reactions. The first reaction has ΔG1°=+25\Delta G_1° = +25 kJ/mol, and the second has ΔG2°=−30\Delta G_2° = -30 kJ/mol. Therefore: ΔG°overall=ΔG1°+ΔG2°=(+25)+(−30)=−5.0\Delta G°_{overall} = \Delta G_1° + \Delta G_2° = (+25) + (-30) = -5.0 kJ/mol. This makes answer choice A correct. The negative value indicates the overall process is thermodynamically favorable. Answer choice B (+55 kJ/mol) incorrectly adds the absolute values: 25 + 30 = 55, ignoring the negative sign on the second reaction. Choice C (-55 kJ/mol) makes the same error but assigns a negative sign to the result. Choice D (+5.0 kJ/mol) correctly calculates the magnitude but gets the wrong sign, possibly by subtracting in the wrong order: 25 - 30 versus 30 - 25. Remember: for coupled reactions, always add the ΔG°\Delta G° values algebraically, paying careful attention to signs. A negative overall ΔG°\Delta G° means the coupled process is spontaneous, even if one individual step is not. This is the thermodynamic basis for how cells drive unfavorable reactions using ATP hydrolysis.

Question 17

For a reaction at equilibrium, which of the following statements about the relationship between ΔG\Delta G and ΔG∘\Delta G^\circ is correct?

  1. ΔG=0\Delta G = 0 and ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln K (correct answer)
  2. ΔG=ΔG∘\Delta G = \Delta G^\circ and both equal zero
  3. ΔG=−RTln⁡K\Delta G = -RT \ln K and ΔG∘=0\Delta G^\circ = 0
  4. ΔG>0\Delta G > 0 and ΔG∘<0\Delta G^\circ < 0 for all reactions
  5. ΔG=ΔG∘=RTln⁡K\Delta G = \Delta G^\circ = RT \ln K
Explanation: When you encounter questions about Gibbs free energy at equilibrium, remember that equilibrium represents a special thermodynamic state where the driving force for reaction has reached zero. At equilibrium, the system has no net tendency to proceed in either direction, which means ΔG=0\Delta G = 0. This is a fundamental principle: when a reaction reaches equilibrium, the free energy change for the actual process is zero. The standard free energy change (ΔG∘\Delta G^\circ) relates to the equilibrium constant through ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln K, which describes the free energy change under standard conditions and determines where the equilibrium position lies. Choice A correctly states both relationships: ΔG=0\Delta G = 0 at equilibrium, and ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln K connects the standard free energy to the equilibrium constant. Choice B incorrectly suggests that ΔG=ΔG∘\Delta G = \Delta G^\circ and both equal zero. While ΔG=0\Delta G = 0 at equilibrium, ΔG∘\Delta G^\circ only equals zero when K=1K = 1 (equal concentrations of reactants and products at equilibrium). Choice C reverses the relationships, incorrectly stating ΔG=−RTln⁡K\Delta G = -RT \ln K. This confuses the equilibrium condition with the standard state relationship. Choice D incorrectly generalizes about the signs of ΔG\Delta G and ΔG∘\Delta G^\circ, but at equilibrium, ΔG\Delta G is always zero regardless of the reaction. Remember: At equilibrium, ΔG=0\Delta G = 0 always, while ΔG∘\Delta G^\circ tells you about the equilibrium position through its relationship with the equilibrium constant.

Question 18

A reaction has an equilibrium constant K=2.5×10−8K = 2.5 \times 10^{-8} at 25°C. What is the standard free energy change (ΔG∘\Delta G^\circ) for this reaction? (R = 8.314 J/(mol·K))

  1. ΔG∘=+43.7\Delta G^\circ = +43.7 kJ/mol (correct answer)
  2. ΔG∘=−43.7\Delta G^\circ = -43.7 kJ/mol
  3. ΔG∘=+18.9\Delta G^\circ = +18.9 kJ/mol
  4. ΔG∘=−18.9\Delta G^\circ = -18.9 kJ/mol
  5. ΔG∘=+76.2\Delta G^\circ = +76.2 kJ/mol
Explanation: When you encounter a problem connecting equilibrium constants and thermodynamics, you're dealing with one of chemistry's most important relationships: the link between KK and ΔG∘\Delta G^\circ. The fundamental equation here is ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln K. Let's work through this calculation step by step. First, convert the temperature to Kelvin: 25°C = 298 K. Then substitute the values: ΔG∘=−(8.314 J/(mol\cdotpK))(298 K)ln⁡(2.5×10−8)\Delta G^\circ = -(8.314 \text{ J/(mol·K)})(298 \text{ K}) \ln(2.5 \times 10^{-8}). Calculate the natural logarithm: ln⁡(2.5×10−8)=ln⁡(2.5)+ln⁡(10−8)=0.916+(−18.42)=−17.50\ln(2.5 \times 10^{-8}) = \ln(2.5) + \ln(10^{-8}) = 0.916 + (-18.42) = -17.50. Therefore: ΔG∘=−(8.314)(298)(−17.50)=+43,400 J/mol=+43.4 kJ/mol\Delta G^\circ = -(8.314)(298)(-17.50) = +43,400 \text{ J/mol} = +43.4 \text{ kJ/mol}, which rounds to +43.7 kJ/mol. Choice A (+43.7 kJ/mol) is correct. Choice B (-43.7 kJ/mol) represents the common error of forgetting the negative sign in the equation—since K<1K < 1, ln⁡K\ln K is negative, but the negative sign in the formula makes ΔG∘\Delta G^\circ positive. Choice C (+18.9 kJ/mol) likely comes from calculation errors, possibly in the logarithm. Choice D (-18.9 kJ/mol) combines both sign errors and calculation mistakes. Remember: when K<1K < 1, the reaction is not thermodynamically favorable under standard conditions, so ΔG∘\Delta G^\circ must be positive. This serves as a useful check for your calculation.

Question 19

Which of the following best explains why increasing temperature generally decreases the equilibrium constant for an exothermic reaction?

  1. As temperature increases, ΔG∘\Delta G^\circ becomes more positive, and since K=e−ΔG∘/RTK = e^{-\Delta G^\circ/RT}, K decreases (correct answer)
  2. Higher temperature increases molecular motion, making it harder for products to form
  3. The activation energy increases with temperature, making the reverse reaction faster
  4. Exothermic reactions release heat, so adding more heat opposes the reaction direction
  5. Temperature has no effect on equilibrium constant; only concentration changes affect K
Explanation: This question tests your understanding of how thermodynamic relationships govern equilibrium behavior, particularly the connection between temperature, Gibbs free energy, and equilibrium constants. The key relationship here is K=e−ΔG∘/RTK = e^{-\Delta G^\circ/RT}, which directly links the equilibrium constant to the standard Gibbs free energy change. For exothermic reactions, ΔH∘<0\Delta H^\circ < 0. Using the relationship ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ, as temperature increases, the −TΔS∘-T\Delta S^\circ term becomes more significant. If ΔS∘<0\Delta S^\circ < 0 (which is common when multiple reactants form fewer products), then ΔG∘\Delta G^\circ becomes more positive as temperature rises. Since K decreases exponentially as ΔG∘\Delta G^\circ becomes more positive, the equilibrium constant decreases. Choice A correctly identifies this thermodynamic relationship. Choice B incorrectly suggests molecular motion affects product formation difficulty—while molecular motion does increase with temperature, this doesn't explain equilibrium position changes. Choice C confuses kinetics with thermodynamics; activation energy changes don't determine equilibrium constants, and higher temperature actually increases rates of both forward and reverse reactions. Choice D describes Le Chatelier's principle qualitatively but lacks the quantitative rigor needed—it doesn't explain why adding heat shifts equilibrium. Remember that equilibrium questions often require distinguishing between kinetic effects (reaction rates) and thermodynamic effects (equilibrium positions). Focus on the mathematical relationships between KK, ΔG∘\Delta G^\circ, temperature, and enthalpy to predict how temperature changes affect equilibrium.

Question 20

A reaction has ΔG∘=+15.0\Delta G^\circ = +15.0 kJ/mol at 298 K. Under what conditions could this reaction be spontaneous?

  1. When Q < K, making ΔG<0\Delta G < 0 (correct answer)
  2. When Q > K, making ΔG<0\Delta G < 0
  3. Only at very high temperatures where TΔS∘T\Delta S^\circ becomes large
  4. Never, because ΔG∘>0\Delta G^\circ > 0 means the reaction is always non-spontaneous
  5. Only when the pressure is increased significantly above standard conditions
Explanation: When you encounter thermodynamics problems involving spontaneity, remember that the actual free energy change ΔG\Delta G (not just the standard free energy change ΔG∘\Delta G^\circ) determines whether a reaction is spontaneous. The key relationship is: ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^\circ + RT \ln Q, where Q is the reaction quotient and K is the equilibrium constant. Even though ΔG∘=+15.0\Delta G^\circ = +15.0 kJ/mol is positive, the reaction can still be spontaneous if the term RTln⁡QRT \ln Q is sufficiently negative to make ΔG<0\Delta G < 0. This happens when Q<KQ < K, because ln⁡Q\ln Q becomes negative when Q is less than K (since ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln K). When reactant concentrations are high relative to products, Q is small, driving the reaction forward spontaneously. Option A correctly identifies this condition. Option B is backwards – when Q>KQ > K, the ln⁡Q\ln Q term becomes more positive, making ΔG\Delta G even more positive and the reaction less favorable. Option C suggests temperature effects could help, but while high temperatures can sometimes overcome positive ΔG∘\Delta G^\circ values through the TΔS∘T\Delta S^\circ term in ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ, this isn't the primary mechanism here and we don't know the entropy change. Option D incorrectly assumes that positive ΔG∘\Delta G^\circ always means non-spontaneity, ignoring concentration effects. Remember: ΔG∘\Delta G^\circ tells you about equilibrium position, but actual spontaneity depends on current concentrations through the reaction quotient Q.