Gibbs Free Energy & Entropy practice questions

Question 1

A student adds a piece of solid zinc to an aqueous solution of hydrochloric acid, as represented by the balanced equation below:

$$\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)$$

Which of the following best predicts and justifies the sign of the standard entropy change, $\Delta S^\circ$, for this reaction?

Accepted Answer

$\Delta S^\circ > 0$, because a gas is produced from solid and aqueous reactants, leading to a net increase in matter dispersal.

Answer

$\Delta S^\circ > 0$, because the reaction is exothermic and transfers heat energy to the surroundings.

Answer

$\Delta S^\circ < 0$, because the total number of moles of products is less than the total number of moles of reactants.

Answer

$\Delta S^\circ < 0$, because the formation of aqueous zinc ions restricts the motion of surrounding water molecules.

Question 2

The synthesis of ammonia is represented by the following balanced equation:

$\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}$

Which of the following best predicts the sign of the standard entropy change, $\Delta S^\circ$, for the reaction and provides the correct justification?

Accepted Answer

$\Delta S^\circ < 0$, because the reaction results in a net decrease in the number of moles of gas.

Answer

$\Delta S^\circ < 0$, because energy is released as new chemical bonds form in the products.

Answer

$\Delta S^\circ > 0$, because two different element species combine to form a compound with greater structural complexity.

Answer

$\Delta S^\circ > 0$, because the total mass and total number of atoms are conserved during the reaction.

Question 3

A student sets up an experiment using two rigid glass bulbs of equal volume connected by a valve, as represented in the particulate diagram. Initially, Bulb 1 contains a gas sample at a constant temperature of 298 K, and Bulb 2 is evacuated. After the valve is opened, the gas expands isothermally until it is evenly distributed across both bulbs. Which of the following best explains why $\Delta S > 0$ for this process based on the particulate representation?

Accepted Answer

The gas particles occupy a larger total volume, which increases the number of possible spatial arrangements (microstates) available to the system.

Answer

The average kinetic energy of the gas particles increases as they enter Bulb $2$, leading to greater thermal disorder in the system.

Answer

The total number of gas particles in the system increases when the valve is opened, creating a greater quantity of matter.

Answer

The attractive forces between the gas particles become stronger as the gas expands, causing a spontaneous increase in energy.

Question 4

A student investigates whether copper metal can provide the electrons needed to recover nickel metal from a solution. The proposed reaction is

$\text{Ni}^{2+}\text{(aq)}+\text{Cu(s)}\rightarrow\text{Ni(s)}+\text{Cu}^{2+}\text{(aq)}$

The relevant standard reduction potentials are shown below.

| Reduction half-reaction | $E^\circ$ |
|---|---:|
| $\text{Ni}^{2+}\text{(aq)}+2e^-\rightarrow\text{Ni(s)}$ | $-0.25\text{ V}$ |
| $\text{Cu}^{2+}\text{(aq)}+2e^-\rightarrow\text{Cu(s)}$ | $+0.34\text{ V}$ |

The experiment is conducted at $25^\circ\text{C}$ with both aqueous ions initially at $1.0\text{ M}$. Which choice correctly indicates whether the proposed reaction is thermodynamically favorable under standard conditions and whether an external voltage source is required to sustain the reaction as written?

Accepted Answer

Thermodynamically favorable: No; external voltage required: Yes

Answer

Thermodynamically favorable: Yes; external voltage required: No

Answer

Thermodynamically favorable: Yes; external voltage required: Yes

Answer

Thermodynamically favorable: No; external voltage required: No

Question 5

A student evaluating a carbon-capture solvent system compares the following substances at $298\text{ K}$. The listed phase is used for each substance.

| Substance | Formula and phase | Molar mass ($\text{g/mol}$) |
|-----------|-------------------|-------------------------------|
| Carbon dioxide | $\text{CO}_2\text{(g)}$ | $44$ |
| Water | $\text{H}_2\text{O(l)}$ | $18$ |
| Methanol | $\text{CH}_3\text{OH(l)}$ | $32$ |
| Ethanol | $\text{C}_2\text{H}_5\text{OH(l)}$ | $46$ |

Which of the following ranks the substances from greatest to least standard molar entropy, $S^\circ$, with the best justification?

Accepted Answer

$\text{CO}_2\text{(g)} > \text{C}_2\text{H}_5\text{OH(l)} > \text{CH}_3\text{OH(l)} > \text{H}_2\text{O(l)}$, because a gas has greater positional freedom than a liquid, and molecular complexity generally increases the number of accessible microstates within a phase.

Answer

$\text{CO}_2\text{(g)} > \text{CH}_3\text{OH(l)} > \text{C}_2\text{H}_5\text{OH(l)} > \text{H}_2\text{O(l)}$, because the gas has the greatest entropy and, among the two liquid alcohols, lower molar mass allows more accessible motion.

Answer

$\text{CO}_2\text{(g)} > \text{H}_2\text{O(l)} > \text{CH}_3\text{OH(l)} > \text{C}_2\text{H}_5\text{OH(l)}$, because a greater ability to form hydrogen bonds increases the number of accessible molecular arrangements in a liquid.

Answer

$\text{C}_2\text{H}_5\text{OH(l)} > \text{CH}_3\text{OH(l)} > \text{H}_2\text{O(l)} > \text{CO}_2\text{(g)}$, because intermolecular contacts in liquids provide more accessible arrangements than the free motion of particles in a gas.

Question 6

A student evaluates the industrial extraction of iron metal from iron(III) oxide using carbon monoxide gas at $298 \text{ K}$. The thermochemical equations for the individual processes are shown below:

$\text{Fe}_2\text{O}_3(s) \rightarrow 2\,\text{Fe}(s) + \frac{3}{2}\,\text{O}_2(g) \quad \Delta G^\circ = +740 \text{ kJ/mol}_{rxn}$

$\text{CO}(g) + \frac{1}{2}\,\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta G^\circ = -280 \text{ kJ/mol}_{rxn}$

Based on the information above, what is the standard Gibbs free energy change, $\Delta G^\circ$, for the net coupled reaction shown below?

$\text{Fe}_2\text{O}_3(s) + 3\,\text{CO}(g) \rightarrow 2\,\text{Fe}(s) + 3\,\text{CO}_2(g)$

Accepted Answer

$-100 \text{ kJ/mol}_{rxn}$

Answer

$+100 \text{ kJ/mol}_{rxn}$

Answer

$+460 \text{ kJ/mol}_{rxn}$

Answer

$+1020 \text{ kJ/mol}_{rxn}$

Question 7

At $300 \text{ K}$, a chemical reaction has a standard Gibbs free energy change of $\Delta G^\circ = -10.0 \text{ kJ/mol}$. At this temperature, the product $RT$ is equal to $2.5 \text{ kJ/mol}$. Which of the following expressions is equal to the standard equilibrium constant, $K$, for the reaction at $300 \text{ K}$?

Accepted Answer

$e^{4}$

Answer

$e^{-4}$

Answer

$e^{0.25}$

Answer

$e^{10}$

Question 8

A student investigates the decomposition reaction represented below at $300. \text{ K}$.

$2\text{N}_2\text{O}_5\text{(g)} \rightarrow 4\text{NO}_2\text{(g)} + \text{O}_2\text{(g)}$

The thermodynamic parameters for the reaction at $300. \text{ K}$ are given in the table below.

| Thermodynamic Parameter | Value |
|---|---|
| $\Delta H^\circ$ | $-40.0 \text{ kJ/mol}_{rxn}$ |
| $\Delta S^\circ$ | $+200. \text{ J/(mol}_{rxn}\cdot\text{K)}$ |

Based on the data, what is the standard Gibbs free energy change, $\Delta G^\circ$, for the reaction at $300. \text{ K}$?

Accepted Answer

$-100. \text{ kJ/mol}_{rxn}$

Answer

$-40.2 \text{ kJ/mol}_{rxn}$

Answer

$-20.0 \text{ kJ/mol}_{rxn}$

Answer

$+20.0 \text{ kJ/mol}_{rxn}$

Question 9

A student plots the standard Gibbs free energy change, $\Delta G^\circ$, as a function of temperature, $T$, for a chemical reaction, as shown in the graph below. Based on the graph, which of the following claims about the reaction is correct?

Accepted Answer

The reaction is endothermic ($\Delta H^\circ > 0$) and becomes thermodynamically favorable at temperatures above $300 \text{ K}$ because $T\Delta S^\circ$ exceeds $\Delta H^\circ$.

Answer

The reaction is endothermic ($\Delta H^\circ > 0$) and becomes thermodynamically favorable at temperatures above $300 \text{ K}$ because $\Delta S^\circ < 0$.

Answer

The reaction is exothermic ($\Delta H^\circ < 0$) and is thermodynamically favorable at all temperatures below $300 \text{ K}$ because $\Delta S^\circ > 0$.

Answer

The reaction is exothermic ($\Delta H^\circ < 0$) and becomes thermodynamically unfavorable at temperatures above $300 \text{ K}$ because $T\Delta S^\circ$ exceeds $\Delta H^\circ$.

Question 10

A student evaluates the thermal decomposition of sodium hydrogen carbonate in a gas-generating cartridge. The reaction at $298\text{ K}$ is represented by the following equation.

$$2\text{NaHCO}_3\text{(s)} \rightarrow \text{Na}_2\text{CO}_3\text{(s)}+\text{CO}_2\text{(g)}+\text{H}_2\text{O(g)}$$

| Substance | $S^\circ$ $(\text{J mol}^{-1}\text{K}^{-1})$ |
|---|---:|
| $\text{NaHCO}_3\text{(s)}$ | 102.1 |
| $\text{Na}_2\text{CO}_3\text{(s)}$ | 135.0 |
| $\text{CO}_2\text{(g)}$ | 213.8 |
| $\text{H}_2\text{O(g)}$ | 188.8 |

What is the value of $\Delta S^\circ$ for the reaction as written?

Accepted Answer

$333.4\text{ J mol}^{-1}\text{K}^{-1}$

Answer

$-333.4\text{ J mol}^{-1}\text{K}^{-1}$

Answer

$166.7\text{ J mol}^{-1}\text{K}^{-1}$

Answer

$198.4\text{ J mol}^{-1}\text{K}^{-1}$

Question 11

A researcher compares two hypothetical galvanic cells used in metal recovery. Both cells operate at the same temperature and transfer the same number of moles of electrons per mole of reaction. The standard cell potentials satisfy $E_2^\circ=2E_1^\circ$, where $E_1^\circ>0$. The corresponding equilibrium constants are $K_1$ and $K_2$.

The relationships $\Delta G^\circ=-nFE^\circ$ and $\Delta G^\circ=-RT\ln K$ apply. Which of the following correctly relates $K_2$ to $K_1$?

Accepted Answer

$K_2=(K_1)^2$

Answer

$K_2=2K_1$

Answer

$K_2=\sqrt{K_1}$

Answer

$K_2=\dfrac{1}{(K_1)^2}$

Question 12

A materials scientist considers the conversion of diamond to graphite at standard conditions.

$\text{C(s, diamond)} \rightarrow \text{C(s, graphite)}$

| Substance | $\Delta_{\mathrm f}G^\circ$ at $298\text{ K}$ |
|---|---:|
| $\text{C(s, diamond)}$ | $+2.9\text{ kJ/mol}$ |
| $\text{C(s, graphite)}$ | $0.0\text{ kJ/mol}$ |

After several years at $298\text{ K}$, no measurable graphite is detected in a sample of diamond. Which statement best reconciles the thermodynamic data with the observation?

Accepted Answer

The conversion is thermodynamically favorable but effectively unobservable, because graphite has a lower standard Gibbs free energy, whereas the extensive bond rearrangement requires crossing a large activation-energy barrier.

Answer

The conversion is thermodynamically favorable but effectively unobservable, because the product-favored equilibrium constant makes the forward reaction rate extremely small.

Answer

The conversion is thermodynamically unfavorable and effectively unobservable, because the large activation energy raises $\Delta G^\circ$ above zero.

Answer

The conversion is thermodynamically unfavorable and effectively unobservable, because subtracting the product value from the reactant value gives $\Delta G^\circ=+2.9\text{ kJ/mol}$.

Question 13

A student studies the phosphorylation of glucose in a buffered metabolic system. In the simplified equations, $\text{G}$ represents glucose, $\text{G6P}$ represents glucose-6-phosphate, and $\text{P}_{i}$ represents inorganic phosphate. Charge-balancing proton transfers with the buffer are omitted.

| Process | Reaction | $\Delta G^\circ$ |
|---|---|---|
| Glucose phosphorylation | $\text{G(aq)} + \text{P}_{i}\text{(aq)} \rightarrow \text{G6P(aq)} + \text{H}_2\text{O(l)}$ | $+14 \, \text{kJ/mol}$ |
| ATP hydrolysis | $\text{ATP(aq)} + \text{H}_2\text{O(l)} \rightarrow \text{ADP(aq)} + \text{P}_{i}\text{(aq)}$ | $-30 \, \text{kJ/mol}$ |

Which proposed sequence correctly couples ATP hydrolysis to glucose phosphorylation and gives a thermodynamically favorable net process?

Accepted Answer

Sequence:$$
\text{ATP(aq)} + \text{H}_2\text{O(l)} \rightarrow \text{ADP(aq)} + \text{P}_{i}\text{(aq)}
$$$$
\text{G(aq)} + \text{P}_{i}\text{(aq)} \rightarrow \text{G6P(aq)} + \text{H}_2\text{O(l)}
$$Net:$$
\text{ATP(aq)} + \text{G(aq)} \rightarrow \text{ADP(aq)} + \text{G6P(aq)}
$$$$
\Delta G^\circ = -16 \, \text{kJ/mol}
$$

Answer

Sequence: $\text{ATP}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{ADP}(\text{aq}) + \text{P}_{i}(\text{aq})$; $\text{G6P}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{G}(\text{aq}) + \text{P}_{i}(\text{aq})$. Net $\Delta G^\circ = -44 \, \text{kJ/mol}$.

Answer

Sequence:$$
\text{ADP(aq)} + \text{P}_{i}\text{(aq)} \rightarrow \text{ATP(aq)} + \text{H}_2\text{O(l)}
$$$$
\text{G(aq)} + \text{P}_{i}\text{(aq)} \rightarrow \text{G6P(aq)} + \text{H}_2\text{O(l)}
$$Net $\Delta G^\circ = +44 \, \text{kJ/mol}$.

Answer

Sequence: $\text{ATP(aq)} + \text{H}_2\text{O(l)} \rightarrow \text{ADP(aq)} + \text{P}_{i}\text{(aq)}$; $\text{G(aq)} + \text{P}_{i}\text{(aq)} \rightarrow \text{G6P(aq)} + \text{H}_2\text{O(l)}$.Net: $\text{ATP(aq)} + \text{G(aq)} \rightarrow \text{ADP(aq)} + \text{G6P(aq)}$, $\Delta G^\circ = +16 \,\text{kJ/mol}$.

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