Free Energy and Equilibrium - College Chemistry

1. The equilibrium constant for the reaction $A + B \rightleftharpoons C + D$ is $K_1 = 4.0 \times 10^3$ at 298 K. What is the equilibrium constant for the reaction $2C + 2D \rightleftharpoons 2A + 2B$ at the same temperature?
   Answer: $K_2 = 6.25 \times 10^{-8}$ 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 $A + B \rightleftharpoons C + D$, we have $K_1 = 4.0 \times 10^3$. The target reaction $2C + 2D \rightleftharpoons 2A + 2B$ 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 $A + B \rightleftharpoons C + D$ to get $C + D \rightleftharpoons A + B$ gives us $K = \frac{1}{4.0 \times 10^3} = 2.5 \times 10^{-4}$. Second, multiplying all coefficients by 2 means raising the equilibrium constant to the power of 2. Therefore: $K_2 = (2.5 \times 10^{-4})^2 = 6.25 \times 10^{-8}$. This confirms answer A is correct. Looking at the wrong answers: B gives $2.5 \times 10^{-4}$, which is just the reverse reaction without accounting for doubling the coefficients. C gives $1.6 \times 10^7$, which incorrectly doubles the original constant instead of reversing it. D gives $8.0 \times 10^3$, 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 *n*th power. Always apply these operations in sequence when manipulating equilibrium expressions.
2. A biochemical reaction has $\Delta G^\circ = +8.5$ kJ/mol. In order for this reaction to be spontaneous in a living cell, which condition must be met?
   Answer: The concentration of reactants must be kept much higher than products to make Q < K When you encounter thermodynamics problems involving biological systems, remember that spontaneity depends on the actual free energy change ($\Delta G$), not just the standard free energy change ($\Delta G^\circ$). The relationship is: $\Delta G = \Delta G^\circ + RT\ln Q$, where Q is the reaction quotient. Since $\Delta G^\circ = +8.5$ kJ/mol is positive, this reaction is non-spontaneous under standard conditions. For spontaneity, you need $\Delta G < 0$. This means the term $RT\ln Q$ must be sufficiently negative to overcome the positive $\Delta G^\circ$. Since $RT$ is always positive, $\ln Q$ must be negative, which occurs when $Q < 1$. 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 $Q < K$ (since $K = e^{-\Delta G^\circ/RT}$), 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 $\Delta G^\circ$ 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 ($Q > K$) would make the reaction even less favorable. Remember: for endergonic reactions ($\Delta G^\circ > 0$) to proceed spontaneously, living systems either couple them to exergonic processes or manipulate concentrations to make the reaction quotient very small.
3. The data in the table shows how the equilibrium constant varies with temperature for the reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$. Based on this data, what can be concluded about $\Delta H^\circ$ for this reaction?
   Answer: $\Delta H^\circ > 0$ because K increases with increasing temperature When the equilibrium constant increases with increasing temperature, this indicates an endothermic reaction ($\Delta H^\circ > 0$). According to Le Châtelier's principle, adding heat (increasing T) favors the endothermic direction, increasing K. The van't Hoff equation also shows that for endothermic reactions, K increases with T. Choice B has the wrong sign relationship. Choice C incorrectly relates linearity to $\Delta H^\circ = 0$. Choice D is wrong because the K vs. T relationship directly indicates the sign of $\Delta H^\circ$. Choice E gives correct answer but wrong reasoning.
4. For a reaction at 298 K, $\Delta H^\circ = -85.0$ kJ/mol and $\Delta S^\circ = -125$ J/(mol·K). What is the value of $\Delta G^\circ$ and what does this indicate about the spontaneity of the reaction under standard conditions?
   Answer: $\Delta G^\circ = -47.8$ kJ/mol; the reaction is spontaneous When you encounter thermodynamics problems involving spontaneity, you need to calculate the Gibbs free energy change using the fundamental equation: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. 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 $\Delta H^\circ = -85.0$ kJ/mol, $\Delta S^\circ = -125$ J/(mol·K), and $T = 298$ K. First, convert the entropy units to match enthalpy: $\Delta S^\circ = -125$ J/(mol·K) = $-0.125$ kJ/(mol·K). Now substitute: $\Delta G^\circ = -85.0 - (298)(-0.125) = -85.0 + 37.3 = -47.8$ kJ/mol. Since $\Delta G^\circ < 0$, 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 $T\Delta S^\circ$ term instead of subtracting it, treating both terms as negative. Answer D (+37.3 kJ/mol) appears to come from using only the $T\Delta S^\circ$ term while ignoring the enthalpy contribution entirely. Remember this key relationship: negative $\Delta G^\circ$ 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.
5. Which of the following best explains why increasing temperature generally decreases the equilibrium constant for an exothermic reaction?
   Answer: As temperature increases, $\Delta G^\circ$ becomes more positive, and since $K = e^{-\Delta G^\circ/RT}$, K decreases 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 $K = e^{-\Delta G^\circ/RT}$, which directly links the equilibrium constant to the standard Gibbs free energy change. For exothermic reactions, $\Delta H^\circ < 0$. Using the relationship $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, as temperature increases, the $-T\Delta S^\circ$ term becomes more significant. If $\Delta S^\circ < 0$ (which is common when multiple reactants form fewer products), then $\Delta G^\circ$ becomes more positive as temperature rises. Since K decreases exponentially as $\Delta G^\circ$ 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 $K$, $\Delta G^\circ$, temperature, and enthalpy to predict how temperature changes affect equilibrium.
6. The relationship between the equilibrium constant and temperature for the reaction $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ follows the van't Hoff equation. If $K_p = 2.8 \times 10^{12}$ at 298 K and $K_p = 2.5 \times 10^{10}$ at 398 K, what can be concluded about this reaction?
   Answer: The reaction is exothermic because K decreases with increasing temperature 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 $K_p$ responds to temperature changes. Looking at the data, $K_p$ decreases dramatically from $2.8 \times 10^{12}$ at 298 K to $2.5 \times 10^{10}$ 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 $K_p$ 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 $\Delta G°$ more negative - in fact, for exothermic reactions, $\Delta G°$ 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 $K_p$ decreases as temperature increases, the reaction is exothermic. If $K_p$ increases with temperature, it's endothermic. This relationship is fundamental to chemical equilibrium.
7. For the reaction $2NO_2(g) \rightleftharpoons N_2O_4(g)$, $\Delta G^\circ = -4.73$ kJ/mol at 298 K. If the partial pressures are $P_{NO_2} = 0.50$ atm and $P_{N_2O_4} = 2.0$ atm, what is $\Delta G$ for the reaction under these conditions? (R = 8.314 J/(mol·K))
   Answer: $\Delta G = +0.42$ kJ/mol When you encounter a problem asking for $\Delta G$ 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 $\Delta G$ under these conditions, use: $\Delta G = \Delta G^\circ + RT \ln Q$, where $Q$ is the reaction quotient. For this reaction, $Q = \frac{P_{N_2O_4}}{(P_{NO_2})^2} = \frac{2.0}{(0.50)^2} = \frac{2.0}{0.25} = 8.0$. Now calculate: $\Delta G = -4.73 \text{ kJ/mol} + (8.314 \text{ J/(mol}\cdot\text{K)})(298 \text{ K}) \ln(8.0)$ $\Delta G = -4730 \text{ J/mol} + (2477.6)(2.08) = -4730 + 5153 = +423 \text{ J/mol} = +0.42 \text{ kJ/mol}$ This confirms answer A is correct. Answer B ($-4.73$ kJ/mol) incorrectly assumes these are standard conditions, ignoring the $RT \ln Q$ term entirely. Answer C ($-9.88$ kJ/mol) likely results from incorrectly subtracting the $RT \ln Q$ term instead of adding it. Answer D ($+3.43$ kJ/mol) probably comes from a calculation error in either $Q$ or the natural logarithm. Remember: $\Delta G^\circ$ only applies at standard conditions (1 atm for gases). When pressures differ from standard conditions, always use $\Delta G = \Delta G^\circ + RT \ln Q$ to account for the actual reaction conditions.
8. For a reaction where $\Delta G^\circ = +12.5$ kJ/mol, what is the ratio of products to reactants at equilibrium at 298 K? (Assume the reaction is $A \rightleftharpoons B$ with equal stoichiometric coefficients)
   Answer: $\frac{[B]}{[A]} = 0.0067$ 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 $\Delta G^\circ$ directly determines the equilibrium constant through the equation $\Delta G^\circ = -RT \ln K$. Let's solve this step by step. First, rearrange to find K: $K = e^{-\Delta G^\circ/RT}$. Substituting the values ($\Delta G^\circ = +12,500$ J/mol, $R = 8.314$ J/mol·K, $T = 298$ K): $K = e^{-12,500/(8.314 \times 298)} = e^{-5.05} = 0.0067$ For the reaction $A \rightleftharpoons B$, the equilibrium constant equals $\frac{[B]}{[A]}$, 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 $+\Delta G^\circ/RT$ in the exponent instead of $-\Delta G^\circ/RT$. Since $\Delta G^\circ$ 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 $\Delta G^\circ$ is in J/mol when using $R = 8.314$ J/mol·K. Remember: positive $\Delta G^\circ$ means $K < 1$, so reactants are favored. Always double-check your sign and units—these are the most common sources of error in thermodynamic calculations.
9. For a gas-phase reaction at constant temperature, if the total pressure is doubled while keeping the mole fractions constant, how does this affect $\Delta G$ for the reaction $A(g) + B(g) \rightleftharpoons C(g)$?
   Answer: $\Delta G$ becomes more negative because Q decreases When you encounter gas-phase equilibrium problems involving pressure changes, you need to connect the relationship between pressure, reaction quotient Q, and Gibbs free energy through the equation $\Delta G = \Delta G° + RT \ln Q$. For this reaction, the reaction quotient is $Q = \frac{P_C}{P_A \cdot P_B}$. When total pressure doubles while mole fractions stay constant, each partial pressure doubles. So the new Q becomes $Q' = \frac{2P_C}{(2P_A)(2P_B)} = \frac{2P_C}{4P_A P_B} = \frac{1}{2} \cdot \frac{P_C}{P_A P_B} = \frac{Q}{2}$. Since Q decreases by half, $\ln Q'$ becomes more negative than $\ln Q$. Because $\Delta G = \Delta G° + RT \ln Q$, and RT is positive, a more negative $\ln Q$ term makes $\Delta G$ more negative. This means answer A is correct. Looking at the wrong answers: Answer B incorrectly states that Q increases—we showed it actually decreases. Answer C falls into the trap of thinking constant mole fractions mean no change in Q, but this ignores how pressure affects the reaction quotient when there's a net change in moles of gas (2 moles reactants → 1 mole product). Answer D reaches the right conclusion about $\Delta G$ becoming more negative but gives an incorrect reason—it's not about reaction favorability but about the mathematical relationship between pressure and Q. Study tip: For gas equilibria, always check how pressure changes affect Q by looking at the stoichiometry. When moles of gas change in the reaction, doubling pressure won't simply cancel out in the Q expression.
10. A reaction has an equilibrium constant $K = 2.5 \times 10^{-8}$ at 25°C. What is the standard free energy change ($\Delta G^\circ$) for this reaction? (R = 8.314 J/(mol·K))
    Answer: $\Delta G^\circ = +43.7$ kJ/mol When you encounter a problem connecting equilibrium constants and thermodynamics, you're dealing with one of chemistry's most important relationships: the link between $K$ and $\Delta G^\circ$. The fundamental equation here is $\Delta G^\circ = -RT \ln K$. Let's work through this calculation step by step. First, convert the temperature to Kelvin: 25°C = 298 K. Then substitute the values: $\Delta G^\circ = -(8.314 \text{ J/(mol}\cdot\text{K)})(298 \text{ K}) \ln(2.5 \times 10^{-8})$. Calculate the natural logarithm: $\ln(2.5 \times 10^{-8}) = \ln(2.5) + \ln(10^{-8}) = 0.916 + (-18.42) = -17.50$. Therefore: $\Delta G^\circ = -(8.314)(298)(-17.50) = +43,400 \text{ J/mol} = +43.4 \text{ kJ/mol}$, 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 $K < 1$, $\ln K$ is negative, but the negative sign in the formula makes $\Delta G^\circ$ 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 $K < 1$, the reaction is not thermodynamically favorable under standard conditions, so $\Delta G^\circ$ must be positive. This serves as a useful check for your calculation.
11. A reaction has $\Delta G^\circ = -8.5$ kJ/mol at 298 K. Under cellular conditions where the actual concentrations give $Q = 150$, what is $\Delta G$ for this reaction, and is it spontaneous under these conditions?
    Answer: $\Delta G = +3.9$ kJ/mol; the reaction is non-spontaneous When you encounter questions about reaction spontaneity under non-standard conditions, you need to distinguish between $\Delta G^\circ$ (standard conditions) and $\Delta G$ (actual conditions). The key relationship is: $\Delta G = \Delta G^\circ + RT \ln Q$, where $Q$ is the reaction quotient under actual conditions. To find $\Delta G$, substitute the given values: $\Delta G = -8.5 \text{ kJ/mol} + (8.314 \times 10^{-3} \text{ kJ/mol}\cdot\text{K})(298 \text{ K}) \ln(150)$ Calculate: $RT = 2.48 \text{ kJ/mol}$ and $\ln(150) = 5.01$ Therefore: $\Delta G = -8.5 + (2.48)(5.01) = -8.5 + 12.4 = +3.9 \text{ kJ/mol}$ Since $\Delta G > 0$, the reaction is non-spontaneous under these cellular conditions. Answer A is correct with $\Delta G = +3.9$ kJ/mol and non-spontaneous classification. Answer B incorrectly subtracts the $RT \ln Q$ term instead of adding it, giving a negative value. Answer C ignores the concentration effects entirely, using only $\Delta G^\circ$. Answer D has the right sign and spontaneity conclusion but contains a calculation error in the magnitude. Study tip: Remember that high product concentrations ($Q > 1$) make reactions less favorable than standard conditions predict. Always check whether $Q$ is greater or less than 1 to anticipate if $\Delta G$ will be more or less negative than $\Delta G^\circ$.
12. For the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, $K_p = 6.0 \times 10^{-2}$ at 500°C. What is $\Delta G^\circ$ for this reaction at 500°C? (R = 8.314 J/(mol·K))
    Answer: $\Delta G^\circ = +18.2$ kJ/mol When you encounter equilibrium problems asking for thermodynamic quantities, you need to connect the equilibrium constant to Gibbs free energy using the fundamental relationship: $\Delta G° = -RT \ln K_p$. First, convert the temperature to Kelvin: 500°C + 273 = 773 K. Then substitute the given values into the equation: $\Delta G° = -RT \ln K_p = -(8.314 \text{ J/mol}\cdot\text{K})(773 \text{ K}) \ln(6.0 \times 10^{-2})$ Calculate the natural logarithm: $\ln(6.0 \times 10^{-2}) = \ln(0.060) = -2.81$ $\Delta G° = -(8.314)(773)(-2.81) = +18,100 \text{ J/mol} = +18.1 \text{ kJ/mol}$ 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 $K_p < 1$, $\ln K_p$ is negative, but the negative sign in the equation makes $\Delta G°$ 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 $K_p < 1$, the reaction favors reactants at equilibrium, so $\Delta G° > 0$ (nonspontaneous under standard conditions). Always double-check your temperature conversion to Kelvin and be careful with signs when $K_p < 1$.
13. The standard free energy change for the formation of glucose from CO₂ and H₂O is $\Delta G^\circ = +2870$ 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)
    Answer: 98 photons 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 $E = \frac{hc}{\lambda}$: $E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{680 \times 10^{-9}} = 2.92 \times 10^{-19} \text{ J per photon}$ Since glucose formation requires +2870 kJ/mol, you need this much energy per molecule: $\frac{2870 \times 10^3 \text{ J/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} = 4.77 \times 10^{-18} \text{ J per molecule}$ The minimum number of photons needed is: $\frac{4.77 \times 10^{-18}}{2.92 \times 10^{-19}} = 16.3 \text{ photons}$ 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.
14. At what temperature will a reaction with $\Delta H^\circ = +65.0$ kJ/mol and $\Delta S^\circ = +180$ J/(mol·K) become spontaneous under standard conditions?
    Answer: T > 361 K When you encounter thermodynamics problems asking about spontaneity, you need to use the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$. A reaction becomes spontaneous when $\Delta G < 0$, which means we need to find the temperature where $\Delta G = 0$ (the threshold). Setting up the equation at the threshold: $0 = \Delta H - T\Delta S$, so $T = \frac{\Delta H}{\Delta S}$. First, convert units so they match. The enthalpy is given in kJ/mol, so convert: $\Delta H = 65.0 \text{ kJ/mol} = 65,000 \text{ J/mol}$. Now calculate: $T = \frac{65,000 \text{ J/mol}}{180 \text{ J/(mol}\cdot\text{K)}} = 361 \text{ K}$ Since both $\Delta H$ and $\Delta S$ are positive, this is an endothermic reaction that increases entropy. At low temperatures, the positive $\Delta H$ term dominates and $\Delta G > 0$ (nonspontaneous). At high temperatures, the $-T\Delta S$ term becomes large enough to make $\Delta G < 0$ (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 $\Delta S$ 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 $\Delta H$ with positive $\Delta S$ means spontaneity increases with temperature.
15. For the coupled reaction: $A + B \rightleftharpoons C + D$ ($\Delta G_1^\circ = +25$ kJ/mol) and $D + E \rightleftharpoons F + G$ ($\Delta G_2^\circ = -30$ kJ/mol), what is the overall $\Delta G^\circ$ for the net reaction $A + B + E \rightleftharpoons C + F + G$?
    Answer: $\Delta G^\circ = -5.0$ kJ/mol 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 $\Delta G°$ is a state function - it depends only on the initial and final states, not the pathway. To find the overall $\Delta G°$, you simply add the individual free energy changes of the coupled reactions. The first reaction has $\Delta G_1° = +25$ kJ/mol, and the second has $\Delta G_2° = -30$ kJ/mol. Therefore: $\Delta G°_{overall} = \Delta G_1° + \Delta G_2° = (+25) + (-30) = -5.0$ 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 $\Delta G°$ values algebraically, paying careful attention to signs. A negative overall $\Delta G°$ 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.
16. In a galvanic cell, the relationship between the cell potential and the equilibrium constant is given by $\Delta G^\circ = -nFE^\circ = -RT \ln K$. For a cell with $E^\circ = +0.76$ 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))
    Answer: $K = 2.9 \times 10^{25}$ 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 $\Delta G^\circ = -nFE^\circ = -RT \ln K$ connects these concepts directly. To find the equilibrium constant, rearrange the equation to solve for K: $\ln K = \frac{nFE^\circ}{RT}$ Substituting the given values: n = 2 electrons, F = 96,485 C/mol, $E^\circ = +0.76$ V, R = 8.314 J/(mol·K), and T = 298 K (25°C): $\ln K = \frac{(2)(96,485)(0.76)}{(8.314)(298)} = \frac{146,657}{2,477} = 59.2$ Therefore: $K = e^{59.2} = 2.9 \times 10^{25}$ Answer A ($1.2 \times 10^{26}$) represents a calculation error, likely from incorrect temperature conversion or rounding mistakes. Answer B ($3.8 \times 10^{13}$) suggests using n = 1 instead of n = 2, a common oversight when the problem states "two-electron transfer." Answer D ($6.7 \times 10^{12}$) appears to result from using incorrect units or forgetting to convert temperature to Kelvin. The correct answer is C: $K = 2.9 \times 10^{25}$. Remember that large positive $E^\circ$ 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.
17. The free energy change for the hydrolysis of ATP is $\Delta G^\circ = -30.5$ kJ/mol under standard conditions. In a cell where [ATP] = 5.0 mM, [ADP] = 1.0 mM, and [Pi] = 2.0 mM, what is $\Delta G$ for ATP hydrolysis at 37°C? (R = 8.314 J/(mol·K))
    Answer: $\Delta G = -50.7$ kJ/mol 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: $\Delta G = \Delta G^\circ + RT \ln Q$, where Q is the reaction quotient. For ATP hydrolysis (ATP → ADP + Pi), the reaction quotient is $Q = \frac{[ADP][Pi]}{[ATP]}$. Substituting the given concentrations: $Q = \frac{(1.0 \times 10^{-3})(2.0 \times 10^{-3})}{5.0 \times 10^{-3}} = 4.0 \times 10^{-4}$ Now calculate the correction term: $RT \ln Q = (8.314)(310.15) \ln(4.0 \times 10^{-4}) = 2577 \times (-7.82) = -20,152$ J/mol = -20.2 kJ/mol Therefore: $\Delta G = -30.5 + (-20.2) = -50.7$ 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.
18. For the reaction $2A(g) \rightleftharpoons B(g) + C(g)$, $\Delta G^\circ = -5.7$ 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 $\Delta G$ when the partial pressure of B reaches 0.8 atm?
    Answer: $\Delta G = -2.3$ kJ/mol When you encounter equilibrium problems involving both standard and actual conditions, you need to distinguish between $\Delta G^\circ$ (standard conditions) and $\Delta G$ (actual reaction conditions). The key relationship is: $\Delta G = \Delta G^\circ + RT \ln Q$, where $Q$ is the reaction quotient. First, determine the partial pressures when $P_B = 0.8$ atm. From stoichiometry, if 0.8 atm of B forms, then 0.8 atm of C also forms, and $1.6$ atm of A is consumed (since the coefficient of A is 2). This leaves $P_A = 2.0 - 1.6 = 0.4$ atm. Now calculate the reaction quotient: $Q = \frac{P_B \cdot P_C}{P_A^2} = \frac{(0.8)(0.8)}{(0.4)^2} = \frac{0.64}{0.16} = 4$ Apply the equation: $\Delta G = -5700 + (8.314)(298)\ln(4) = -5700 + 2477(1.386) = -5700 + 3433 = -2267$ J/mol $= -2.3$ kJ/mol This confirms answer A is correct. Answer B (-8.1 kJ/mol) might result from incorrectly subtracting the $RT\ln Q$ term instead of adding it. Answer C (+2.4 kJ/mol) could come from sign errors with $\Delta G^\circ$. Answer D (-11.4 kJ/mol) might result from calculation errors in the logarithm or reaction quotient. Remember: $\Delta G$ tells you the driving force under actual conditions, while $\Delta G^\circ$ only applies to standard conditions. Always check your stoichiometry when determining equilibrium partial pressures.
19. A reaction has $\Delta H^\circ = -45$ kJ/mol and $\Delta S^\circ = -135$ J/(mol·K). At what temperature will the equilibrium constant equal 1.0?
    Answer: T = 333 K This question tests your understanding of the relationship between thermodynamics and equilibrium. When the equilibrium constant equals 1.0, the Gibbs free energy change ($\Delta G^\circ$) 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: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. Since $\Delta G^\circ = 0$ when K = 1.0, you can set up: $0 = \Delta H^\circ - T\Delta S^\circ$, which rearranges to $T = \frac{\Delta H^\circ}{\Delta S^\circ}$. First, convert units so they match. Since $\Delta H^\circ = -45$ kJ/mol and $\Delta S^\circ = -135$ J/(mol·K), convert enthalpy to J/mol: $\Delta H^\circ = -45,000$ J/mol. Now calculate: $T = \frac{-45,000 \text{ J/mol}}{-135 \text{ J/(mol}\cdot\text{K)}} = 333 \text{ K}$ 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 $\Delta G^\circ = 0$ 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.
20. Consider a system where the reaction $A + B \rightleftharpoons C + D$ has reached equilibrium. If additional reactant A is added to the system, which statement correctly describes the immediate effect on $\Delta G$ and the system's response?
    Answer: $\Delta G$ becomes negative, and the system shifts right to reestablish equilibrium 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, $\Delta G = 0$ 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 $Q$ smaller than the equilibrium constant $K$. Since $\Delta G = RT \ln(Q/K)$, when $Q < K$, the natural logarithm becomes negative, making $\Delta G$ negative. A negative $\Delta G$ 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: $\Delta G$ becomes negative and the system shifts right. Choice B incorrectly suggests $\Delta G$ becomes positive and the system shifts left—this would happen if you removed reactants or added excess products. Choice C wrongly assumes $\Delta G$ stays zero; while the system will return to $\Delta G = 0$ at the new equilibrium, the disturbance immediately makes $\Delta G$ non-zero. Choice D correctly identifies that $\Delta G$ 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 $\Delta G$ non-zero, and the sign of $\Delta G$ tells you which direction the system will shift to restore equilibrium.

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