College Chemistry Quiz: Gibbs Free Energy And Thermodynamic Favorability

20 questions · exam conditions

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Gibbs Free Energy And Thermodynamic FavorabilityQuestion 1 of 20

If $\Delta G^=0$ for a reaction at 298 K, what is the value of $K$ ?

K=0

, so the reaction cannot proceed in either direction.

K=1

, so neither reactants nor products are favored.

K>1

, so products are strongly favored at equilibrium.

K<1

, so reactants are strongly favored at equilibrium.

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College Chemistry Quiz

College Chemistry Quiz: Gibbs Free Energy And Thermodynamic Favorability

Practice Gibbs Free Energy And Thermodynamic Favorability 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 Gibbs Free Energy And Thermodynamic Favorability, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

How to use this quiz

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.

All questions

Question 1

If $\Delta G^=0$ for a reaction at 298 K, what is the value of $K$ ?

$K=0$ , so the reaction cannot proceed in either direction.
$K=1$ , so neither reactants nor products are favored. (correct answer)
$K>1$ , so products are strongly favored at equilibrium.
$K<1$ , so reactants are strongly favored at equilibrium.

Explanation: This question tests the understanding of the equilibrium constant when ΔG° = 0. Gibbs Free Energy (ΔG°) indicates whether a process is thermodynamically favorable, with negative ΔG° signifying spontaneity. In the given scenario, ΔG° = 0 implies ln K = 0, so K = 1. The correct answer states that K = 1, so neither reactants nor products are favored. A common mistake involves thinking K = 0 or associating it with no reaction. To help students, emphasize that K = 1 means equal concentrations at equilibrium. Encourage practice with phase transitions where ΔG° = 0 at specific conditions.

Question 2

In $\Delta G=\Delta H-T\Delta S$ , what does each term represent for a reaction at constant temperature and pressure?

$\Delta G$ is heat released; $\Delta H$ is work; $T\Delta S$ is activation energy.
$\Delta G$ is free energy change; $\Delta H$ is enthalpy change; $T\Delta S$ is entropy term. (correct answer)
$\Delta G$ is entropy change; $\Delta H$ is temperature; $T\Delta S$ is enthalpy term.
$\Delta G$ is equilibrium constant; $\Delta H$ is reaction quotient; $T\Delta S$ is rate constant.

Explanation: This question tests the understanding of the components in the Gibbs Free Energy equation ΔG = ΔH - TΔS. Gibbs Free Energy (ΔG) indicates whether a process is thermodynamically favorable, with negative ΔG signifying spontaneity. In the given scenario, ΔH represents the enthalpy change, and -TΔS accounts for the entropy contribution at constant temperature. The correct answer identifies ΔG as the free energy change, ΔH as enthalpy change, and TΔS as the entropy term. A common mistake involves swapping the roles of ΔH and TΔS or confusing them with other concepts like activation energy. To help students, emphasize the physical meaning of each term in energy availability. Encourage practice by deriving the equation from first principles to solidify comprehension.

Question 3

At equilibrium for a reversible reaction at constant $T$ and $P$ , what is the value of $\Delta G$ ?

$\Delta G$ is negative because products are forming spontaneously.
$\Delta G$ is zero because forward and reverse driving forces balance. (correct answer)
$\Delta G$ is positive because reactants are favored at equilibrium.
$\Delta G$ is undefined because equilibrium stops all molecular motion.

Explanation: This question tests the understanding of Gibbs Free Energy at equilibrium for reversible reactions. Gibbs Free Energy (ΔG) indicates whether a process is thermodynamically favorable, with negative ΔG signifying spontaneity. In the given scenario, at equilibrium, the forward and reverse rates balance, resulting in ΔG = 0. The correct answer states that ΔG is zero because forward and reverse driving forces balance. A common mistake involves confusing ΔG with ΔG° or thinking equilibrium means positive ΔG. To help students, emphasize that ΔG = 0 defines equilibrium, not cessation of reaction. Encourage practice with reaction quotients to predict shifts from equilibrium.

Question 4

Explain why a reaction with a positive $\Delta G$ can still occur.

It can proceed if coupled to a more negative $\Delta G$ process overall. (correct answer)
Positive $\Delta G$ means the reaction is spontaneous but slow.
A catalyst makes $\Delta G$ negative by lowering $\Delta H$ .
Positive $\Delta G$ means the reaction is already at equilibrium.

Explanation: This question tests the understanding of how reactions with positive ΔG can still proceed. Gibbs Free Energy (ΔG) indicates whether a process is thermodynamically favorable, with negative ΔG signifying spontaneity. In the given scenario, a positive ΔG reaction can occur if coupled to another with a more negative ΔG, making the overall process spontaneous. The correct answer highlights that it can proceed if coupled to a more negative ΔG process. A common mistake involves thinking catalysts change ΔG, but they only affect rates. To help students, emphasize biological examples like ATP coupling. Encourage practice with coupled reaction calculations to understand energy transfer.

Question 5

If $K=2.0\times 10^{3}$ at 298 K, what is the sign of $\Delta G^$ ?

$\Delta G^$ is positive because $K$ is greater than 1.
$\Delta G^$ is negative because $\ln K$ is positive. (correct answer)
$\Delta G^$ is zero because products and reactants are equal.
$\Delta G^$ is undefined because $K$ has no units.

Explanation: This question tests the understanding of the sign of ΔG° when K > 1. Gibbs Free Energy (ΔG°) indicates whether a process is thermodynamically favorable, with negative ΔG° signifying spontaneity. In the given scenario, K = 2.0 × 10^3 > 1 means ln K > 0, so ΔG° < 0. The correct answer states that ΔG° is negative because ln K is positive. A common mistake involves associating large K with positive ΔG°. To help students, emphasize how favorable reactions have large K and negative ΔG°. Encourage practice with acid dissociation constants as examples.

Question 6

A reaction has $\Delta G=0$ under certain conditions; what does this imply about the reaction quotient $Q$ ?

$Q=K$ , so the system is at equilibrium under those conditions. (correct answer)
$Q>K$ , so the forward reaction is spontaneous.
$Q<K$ , so the reverse reaction is spontaneous.
$Q=0$ , so no products are present and reaction must stop.

Explanation: This question tests the understanding of the reaction quotient when ΔG = 0. Gibbs Free Energy (ΔG) indicates whether a process is thermodynamically favorable, with negative ΔG signifying spontaneity. In the given scenario, ΔG = 0 implies the system is at equilibrium, so Q = K. The correct answer states that Q = K, so the system is at equilibrium. A common mistake involves thinking ΔG = 0 means Q = 0 or directional preferences. To help students, emphasize the equation ΔG = ΔG° + RT ln Q. Encourage practice with non-standard conditions to calculate Q and predict shifts.

Question 7

For electrolysis: $2\text{H}_2\text{O}(l) \rightarrow 2\text{H}_2(g)+\text{O}_2(g)$ has $\Delta G^>0$ ; what follows?

The reaction is spontaneous and releases electrical energy to the surroundings.
The reaction requires external electrical work to proceed under standard conditions. (correct answer)
The reaction is at equilibrium because $\Delta G^$ is positive.
The reaction rate is fast because $\Delta G^$ is large.

Explanation: This question tests the understanding of Gibbs Free Energy for nonspontaneous processes like electrolysis. Gibbs Free Energy (ΔG) indicates whether a process is thermodynamically favorable, with negative ΔG signifying spontaneity. In the given scenario, positive ΔG° for water electrolysis means it requires external energy input. The correct answer states that the reaction requires external electrical work to proceed. A common mistake involves assuming positive ΔG means equilibrium or fast rates. To help students, emphasize the role of ΔG in electrochemical cells. Encourage practice with battery vs. electrolysis examples to distinguish spontaneous and driven processes.

Question 8

How does temperature affect spontaneity when $\Delta H<0$ and $\Delta S<0$ in $\Delta G=\Delta H-T\Delta S$ ?

Higher $T$ makes $\Delta G$ more negative, so spontaneity increases.
Higher $T$ makes $\Delta G$ more positive, so spontaneity decreases. (correct answer)
Temperature cannot change $\Delta G$ when both $\Delta H$ and $\Delta S$ are negative.
Higher $T$ always makes reactions spontaneous regardless of signs.

Explanation: This question tests the understanding of temperature's effect on Gibbs Free Energy when ΔH < 0 and ΔS < 0. Gibbs Free Energy (ΔG) indicates whether a process is thermodynamically favorable, with negative ΔG signifying spontaneity. In the given scenario, higher temperature makes the -TΔS term more positive because ΔS is negative, thus making ΔG more positive. The correct answer notes that higher T makes ΔG more positive, decreasing spontaneity. A common mistake involves assuming all exothermic reactions are always spontaneous regardless of temperature. To help students, emphasize that at high temperatures, the entropy term can dominate. Encourage practice with examples like gas condensation to illustrate temperature dependence.

Question 9

A reaction has $\Delta H=-25\ \text{kJ/mol}$ and $\Delta S=+80\ \text{J/(mol\u00b7K)}$ ; which is true for $\Delta G$ ?

$\Delta G$ is positive at all temperatures because $\Delta H$ is negative.
$\Delta G$ is negative at all temperatures because both terms favor spontaneity. (correct answer)
$\Delta G$ is zero at all temperatures because $\Delta S$ is positive.
$\Delta G$ changes sign with temperature because $\Delta H$ and $\Delta S$ oppose.

Explanation: This question tests the understanding of Gibbs Free Energy signs when both ΔH and ΔS favor spontaneity. Gibbs Free Energy (ΔG) indicates whether a process is thermodynamically favorable, with negative ΔG signifying spontaneity. In the given scenario, negative ΔH and positive ΔS ensure ΔG is negative at all temperatures. The correct answer states that ΔG is negative at all temperatures because both terms favor spontaneity. A common mistake involves thinking temperature can reverse the sign when terms align. To help students, emphasize that opposing signs lead to temperature dependence, not aligning ones. Encourage practice with real reactions like combustion to illustrate.

Question 10

For combustion of glucose: $\text{C}_6\text{H}_{12}\text{O}_6(s)+6\text{O}_2(g)\rightarrow 6\text{CO}_2(g)+6\text{H}_2\text{O}(l)$ , $\Delta G^<0$ implies what?

It is nonspontaneous unless driven by an applied voltage.
It is spontaneous under standard conditions and releases usable free energy. (correct answer)
It must be at equilibrium because products and reactants coexist.
It proceeds quickly because $\Delta G^$ determines reaction rate.

Explanation: This question tests the understanding of Gibbs Free Energy for spontaneous metabolic reactions like glucose combustion. Gibbs Free Energy (ΔG°) indicates whether a process is thermodynamically favorable, with negative ΔG° signifying spontaneity. In the given scenario, negative ΔG° means the reaction is spontaneous and releases usable energy. The correct answer states it is spontaneous under standard conditions and releases usable free energy. A common mistake involves confusing thermodynamics with kinetics, thinking it implies fast rates. To help students, emphasize biological energy harvesting from such reactions. Encourage practice with calorimetry data to connect ΔG to real processes.

Question 11

If $K=1.0\times 10^{-5}$ at 298 K, what is the sign of $\Delta G^$ ?

$\Delta G^$ is negative because $K$ is less than 1.
$\Delta G^$ is positive because $\ln K$ is negative. (correct answer)
$\Delta G^$ is zero because $K$ is a constant.
$\Delta G^$ cannot be determined without the reaction rate.

Explanation: This question tests the understanding of the sign of ΔG° from the equilibrium constant K. Gibbs Free Energy (ΔG°) indicates whether a process is thermodynamically favorable, with negative ΔG° signifying spontaneity. In the given scenario, K = 1.0 × 10^{-5} < 1 means ln K < 0, so ΔG° > 0. The correct answer states that ΔG° is positive because ln K is negative. A common mistake involves thinking small K implies negative ΔG°. To help students, emphasize the inverse relationship in the formula. Encourage practice with logarithmic calculations for various K values.

Question 12

What does a negative value of $\Delta G$ indicate about a reaction under the stated conditions?

The reaction is spontaneous in the forward direction under those conditions. (correct answer)
The reaction is at equilibrium and no net reaction occurs.
The reaction rate must be fast because $\Delta G$ is negative.
The reaction is nonspontaneous unless a catalyst is added.

Explanation: This question tests the understanding of Gibbs Free Energy in predicting thermodynamic favorability and spontaneity. Gibbs Free Energy (ΔG) indicates whether a process is thermodynamically favorable, with negative ΔG signifying spontaneity. In the given scenario, a negative ΔG means the reaction proceeds spontaneously in the forward direction under the specified conditions. The correct answer highlights that the reaction is spontaneous in the forward direction. A common mistake involves confusing spontaneity with reaction rate, assuming negative ΔG implies a fast reaction. To help students, emphasize that ΔG predicts direction and favorability, not speed. Encourage practice with examples where ΔG is negative but reactions are slow due to kinetics.

Question 13

A reaction at 25°C has $\Delta G^\circ = +15.3$ kJ/mol. If the reaction quotient $Q = 0.025$ , what is $\Delta G$ for the reaction under these non-standard conditions?

+6.1 kJ/mol (correct answer)
+24.5 kJ/mol
-9.2 kJ/mol
+15.3 kJ/mol
+33.6 kJ/mol

Explanation: When you encounter problems involving both standard and non-standard conditions in thermodynamics, you need to connect the standard Gibbs free energy change (

\Delta G^\circ

) with the actual conditions using the reaction quotient Q. The key relationship is:

\Delta G = \Delta G^\circ + RT \ln Q

Given:

\Delta G^\circ = +15.3

kJ/mol, T = 25°C = 298 K, and Q = 0.025 First, convert R to appropriate units: R = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K) Now calculate:

\Delta G = 15.3 + (0.008314)(298)\ln(0.025)

Since

\ln(0.025) = -3.69

\Delta G = 15.3 + (2.48)(-3.69) = 15.3 - 9.2 = +6.1

kJ/mol Looking at the wrong answers: B (+24.5 kJ/mol) likely comes from incorrectly adding the RT ln Q term instead of recognizing that ln(0.025) is negative. C (-9.2 kJ/mol) represents just the RT ln Q term alone, ignoring

\Delta G^\circ

entirely. D (+15.3 kJ/mol) is simply the standard free energy change, suggesting the student forgot that non-standard conditions require the additional RT ln Q correction. The correct answer is A (+6.1 kJ/mol). Study tip: Remember that when Q < 1, ln Q is negative, making the reaction more thermodynamically favorable than under standard conditions. Always check whether ln Q should be positive or negative based on whether Q is greater or less than 1.

Question 14

Which statement best describes the relationship between $\Delta G^\circ$ and the equilibrium constant $K$ for a chemical reaction?

When $\Delta G^\circ = 0$ , the reaction is at equilibrium and $K = 1$
When $\Delta G^\circ < 0$ , the reaction is thermodynamically favorable and $K > 1$ (correct answer)
When $\Delta G^\circ > 0$ , the reaction proceeds spontaneously and $K < 1$
The magnitude of $K$ is directly proportional to $\Delta G^\circ$
$\Delta G^\circ$ and $K$ are independent quantities with no mathematical relationship

Explanation: When you encounter questions linking thermodynamics to equilibrium, focus on the fundamental relationship:

\Delta G^\circ = -RT \ln K

. This equation connects the standard free energy change to the equilibrium constant and reveals how thermodynamic favorability relates to equilibrium position. The correct answer is B because when

\Delta G^\circ < 0

, the natural logarithm of

K

must be positive (since

-RT

is negative). This means

K > 1

, indicating the reaction is thermodynamically favorable with products favored at equilibrium. Let's examine why the other options are incorrect: Option A incorrectly states that

K = 1

when

\Delta G^\circ = 0

. While

\Delta G^\circ = 0

does indicate equilibrium conditions,

K = 1

only when reactants and products have equal concentrations at equilibrium, which isn't always the case. Option C contains a contradiction. When

\Delta G^\circ > 0

, the reaction is thermodynamically unfavorable and does NOT proceed spontaneously under standard conditions. While

K < 1

is correct for this scenario, the claim about spontaneous proceeding is wrong. Option D misrepresents the mathematical relationship. The connection between

\Delta G^\circ

and

K

is logarithmic, not directly proportional. As

\Delta G^\circ

becomes more negative,

K

increases exponentially. Study tip: Memorize

\Delta G^\circ = -RT \ln K

and remember the sign relationships: negative

\Delta G^\circ

means favorable reaction and

K > 1

; positive

\Delta G^\circ

means unfavorable reaction and

K < 1

Question 15

For the reaction $A + B \rightleftharpoons C + D$ , $\Delta G^\circ = -8.4$ kJ/mol at 25°C. If the initial concentrations are [A] = 0.10 M, [B] = 0.20 M, [C] = 0.50 M, and [D] = 0.30 M, what is the reaction quotient $Q$ and in which direction will the reaction proceed?

$Q = 7.5$ ; reaction proceeds toward products because $Q < K$ (correct answer)
$Q = 0.13$ ; reaction proceeds toward products because $Q < K$
$Q = 7.5$ ; reaction proceeds toward reactants because $Q > K$
$Q = 0.13$ ; reaction proceeds toward reactants because $Q > K$
$Q = 30$ ; reaction is at equilibrium because $Q = K$

Explanation: This question tests your understanding of reaction quotients and equilibrium constants, which determine reaction direction. When you see initial concentrations and a standard free energy change, you need to calculate both the reaction quotient

Q

and equilibrium constant

K

to predict which way the reaction will shift. First, calculate the equilibrium constant using

\Delta G° = -RT \ln K

. Rearranging:

K = e^{-\Delta G°/RT} = e^{-(-8400)/(8.314 \times 298)} = e^{3.39} = 29.7

Next, calculate the reaction quotient using

Q = \frac{[C][D]}{[A][B]} = \frac{(0.50)(0.30)}{(0.10)(0.20)} = \frac{0.15}{0.02} = 7.5

Since

Q = 7.5 < K = 29.7

, the reaction proceeds toward products to reach equilibrium. Answer A correctly identifies

Q = 7.5

and the forward direction because

Q < K

. Answer B has the wrong

Q

value (0.13), which would result from incorrectly inverting the quotient expression. Answer C correctly calculates

Q = 7.5

but incorrectly states the reaction proceeds toward reactants; this represents the common misconception of confusing the relationship between

Q

and

K

. Answer D combines both errors: the inverted

Q

calculation and wrong directional reasoning. Remember: when

Q < K

, reactions shift right (toward products); when

Q > K

, they shift left (toward reactants). Always double-check your

Q

expression—products in numerator, reactants in denominator.

Question 16

A reaction has $K_1 = 2.5 \times 10^{-4}$ at 300 K and $K_2 = 1.8 \times 10^{-3}$ at 350 K. What can be concluded about $\Delta H^\circ$ for this reaction?

$\Delta H^\circ > 0$ because $K$ increases with temperature, indicating an endothermic reaction (correct answer)
$\Delta H^\circ < 0$ because $K$ increases with temperature, indicating an exothermic reaction
$\Delta H^\circ = 0$ because the change in $K$ is proportional to the temperature change
$\Delta H^\circ > 0$ because higher temperature decreases the activation energy barrier
Cannot determine the sign of $\Delta H^\circ$ without knowing $\Delta S^\circ$

Explanation: When you encounter equilibrium constants at different temperatures, you're dealing with the van't Hoff equation, which relates how

K

changes with temperature to the enthalpy change of the reaction. The key insight is observing how the equilibrium constant changes as temperature increases. Here,

K

increases from

2.5 \times 10^{-4}

at 300 K to

1.8 \times 10^{-3}

at 350 K - roughly a 7-fold increase. When

K

increases with temperature, it means higher temperature favors the forward reaction, which only happens when the reaction absorbs heat (endothermic,

\Delta H^\circ > 0

). This follows Le Châtelier's principle: adding heat shifts equilibrium toward the endothermic direction. Choice A correctly identifies that

\Delta H^\circ > 0

because

K

increases with temperature, indicating an endothermic reaction. Choice B has the sign backwards - if

\Delta H^\circ < 0

(exothermic), then

K

would decrease with increasing temperature, not increase. Choice C incorrectly suggests

\Delta H^\circ = 0

. When enthalpy change is zero,

K

doesn't change with temperature at all. The proportional relationship mentioned doesn't determine the sign of

\Delta H^\circ

. Choice D mentions activation energy, which affects reaction rate, not equilibrium position. Activation energy is kinetics; equilibrium constants reflect thermodynamics. Study tip: Remember the pattern - when

K

increases with temperature, the reaction is endothermic (

\Delta H^\circ > 0

). When

K

decreases with temperature, it's exothermic (

\Delta H^\circ < 0

). This relationship appears frequently in thermodynamics problems.

Question 17

For a reaction at equilibrium, which statement about Gibbs free energy is correct?

$\Delta G = 0$ and $\Delta G^\circ = 0$ for the system under all conditions
$\Delta G = 0$ for the equilibrium mixture, but $\Delta G^\circ$ depends on the standard state (correct answer)
$\Delta G^\circ = 0$ for the standard state, but $\Delta G$ depends on actual concentrations
Both $\Delta G$ and $\Delta G^\circ$ must be negative for a spontaneous reaction at equilibrium
$\Delta G > 0$ at equilibrium because no net change occurs in either direction

Explanation: When you encounter questions about equilibrium and Gibbs free energy, remember that

\Delta G

and

\Delta G^\circ

represent fundamentally different conditions and have different meanings at equilibrium. At equilibrium, the system has no driving force for net change in either direction, which means

\Delta G = 0

. This is always true for any equilibrium mixture, regardless of the actual concentrations of reactants and products. However,

\Delta G^\circ

refers to the free energy change when all species are at their standard states (1 M concentrations, 1 atm pressure). This value is fixed for a given reaction at a specific temperature and doesn't become zero just because the reaction reaches equilibrium under different conditions. Choice A is incorrect because

\Delta G^\circ

doesn't equal zero for most reactions—it's determined by the inherent thermodynamic properties of the reactants and products under standard conditions. Choice C reverses the correct relationship; it's

\Delta G

(not

\Delta G^\circ

) that equals zero at equilibrium. Choice D misunderstands equilibrium entirely—at equilibrium, the reaction isn't proceeding spontaneously in either direction, so

\Delta G

must be zero, not negative. The key relationship connecting these concepts is:

\Delta G = \Delta G^\circ + RT \ln Q

. At equilibrium,

\Delta G = 0

and

Q = K_{eq}

, giving us

\Delta G^\circ = -RT \ln K_{eq}

. Remember:

\Delta G = 0

defines equilibrium, while

\Delta G^\circ

determines where that equilibrium lies through the equilibrium constant.

Question 18

The combustion of glucose: $C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l)$ has $\Delta H^\circ = -2800$ kJ/mol and $\Delta S^\circ = +180$ J/(mol·K) at 25°C. What is $\Delta G^\circ$ for this reaction?

-2747 kJ/mol
-2854 kJ/mol (correct answer)
-2800 kJ/mol
-53.6 kJ/mol
+53.6 kJ/mol

Explanation: When you encounter thermodynamics problems involving

\Delta G^\circ

, you're working with the fundamental relationship that determines whether a reaction is spontaneous. The key equation here is the Gibbs free energy equation:

\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ

. To solve this problem, you need to substitute the given values carefully, paying attention to units. Given:

\Delta H^\circ = -2800

kJ/mol,

\Delta S^\circ = +180

J/(mol·K), and T = 25°C = 298 K. First, convert

\Delta S^\circ

to kJ:

180 \text{ J/(mol·K)} = 0.180 \text{ kJ/(mol·K)}

Then calculate:

\Delta G^\circ = -2800 \text{ kJ/mol} - (298 \text{ K})(0.180 \text{ kJ/(mol·K)}) = -2800 - 53.6 = -2854 \text{ kJ/mol}

Answer A (-2747 kJ/mol) results from incorrectly adding the entropy term instead of subtracting it, showing a sign error in applying the Gibbs equation. Answer C (-2800 kJ/mol) completely ignores the entropy contribution, suggesting the student forgot the

T\Delta S^\circ

term entirely. Answer D (-53.6 kJ/mol) represents only the entropy contribution (

T\Delta S^\circ

), indicating the student forgot to include the enthalpy term. Always double-check your unit conversions in thermodynamics problems—mixing kJ and J is a common source of errors. Remember that the entropy term becomes more significant at higher temperatures, making

\Delta G^\circ

more negative when

\Delta S^\circ

is positive.

Question 19

A reaction mixture contains products and reactants such that $Q = K$ for the reaction $2A(g) \rightleftharpoons B(g) + C(g)$ . If the temperature is suddenly increased, and the reaction is endothermic, what happens to $\Delta G$ immediately after the temperature change?

$\Delta G$ becomes positive because $Q > K$ at the new temperature
$\Delta G$ becomes negative because $Q < K$ at the new temperature (correct answer)
$\Delta G$ remains zero because the concentrations haven't changed yet
$\Delta G$ becomes positive because $\Delta G^\circ$ increases with temperature
$\Delta G$ becomes negative because $\Delta G^\circ$ decreases with temperature

Explanation: When you encounter equilibrium problems involving temperature changes, you need to consider how temperature affects both the equilibrium constant K and the relationship between Q and K. Initially, the system is at equilibrium with

Q = K

, so

\Delta G = 0

. When temperature suddenly increases for an endothermic reaction, the equilibrium constant K increases according to Le Chatelier's principle - the system favors the forward reaction to absorb the added heat. However, the concentrations of reactants and products don't change instantaneously, so Q remains the same immediately after the temperature change. This creates a situation where

Q < K

at the new temperature. Since

\Delta G = \Delta G^\circ + RT \ln Q - RT \ln K = RT \ln(Q/K)

, when

Q < K

, we have

Q/K < 1

, making

\ln(Q/K)

negative. Therefore,

\Delta G

becomes negative, indicating the reaction will proceed forward to reach the new equilibrium. Choice A incorrectly states

Q > K

, which would occur if the reaction were exothermic. Choice C wrongly assumes

\Delta G

stays zero - while concentrations haven't changed, K has changed, disrupting the equilibrium. Choice D correctly notes that temperature affects the system but incorrectly focuses on

\Delta G^\circ

rather than the Q/K relationship that determines

\Delta G

. Remember: for endothermic reactions, higher temperature increases K, and immediately after the change,

Q < K

makes

\Delta G

negative, driving the reaction forward.

Question 20

Which combination of $\Delta H^\circ$ and $\Delta S^\circ$ values would result in a reaction that is thermodynamically favorable only at low temperatures?

$\Delta H^\circ < 0$ and $\Delta S^\circ > 0$
$\Delta H^\circ > 0$ and $\Delta S^\circ < 0$
$\Delta H^\circ < 0$ and $\Delta S^\circ < 0$ (correct answer)
$\Delta H^\circ > 0$ and $\Delta S^\circ > 0$
$\Delta H^\circ = 0$ and $\Delta S^\circ < 0$

Explanation: When you encounter questions about temperature-dependent spontaneity, you need to analyze how the Gibbs free energy equation

\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ

behaves at different temperatures. A reaction is thermodynamically favorable when

\Delta G^\circ < 0

. For a reaction to be favorable only at low temperatures, it must become unfavorable as temperature increases. This happens when

\Delta H^\circ < 0

and

\Delta S^\circ < 0

. At low temperatures, the

\Delta H^\circ

term dominates, making

\Delta G^\circ

negative (favorable). However, as temperature rises, the

-T\Delta S^\circ

term becomes increasingly positive (since

\Delta S^\circ

is negative), eventually making

\Delta G^\circ

positive (unfavorable). Choice A (

\Delta H^\circ < 0

\Delta S^\circ > 0

) represents a reaction favorable at all temperatures because both terms make

\Delta G^\circ

negative. Choice B (

\Delta H^\circ > 0

\Delta S^\circ < 0

) gives a reaction that's never favorable since both terms make

\Delta G^\circ

positive. Choice D (

\Delta H^\circ > 0

\Delta S^\circ > 0

) describes a reaction favorable only at high temperatures, where the

-T\Delta S^\circ

term eventually overcomes the positive

\Delta H^\circ

. The correct answer is C. Study tip: Remember the phrase "Hot and Cold" for temperature dependence: when enthalpy and entropy have opposite signs, temperature determines favorability. Negative enthalpy with negative entropy means "favorable when cold."