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Free Energy
Which of the following best explains why the reaction occurs at a significant rate only when the temperature of the reaction mixture is below 150 °C?
The reaction is endothermic.
The reaction is exothermic.
The entropy change (ΔS°) for the reaction is positive.
The Gibbs free energy change becomes positive at temperatures above 150 °C
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CaCO3 (s) → CaO (s) + CO2 (g)
The reaction above has a standard entropy change, ΔS° of +161 J mol-1 K-1. The same reaction has a standard enthalpy change, ΔH°, of +178 kJ mol-1.
Calculate the free energy at 25 °C for the reaction.
-47,800
-3,525
+130
+173
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As a sample of KNO3 (s) is stirred into water at 25°C, the compound dissolves endothermically.
Which of the following best helps to explain why the process is thermodynamically favorable at 25°C?
All endothermic processes are thermodynamically favorable.
Stirring the solution during dissolution adds the energy needed to drive an endothermic process.
Dissolving the salt decreases the enthalpy of the system.
Dissolving the salt increases the entropy of the system.
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When solid ammonium chloride dissolves in distilled water, the temperature of the solution decreases.
What are the signs of ΔH°, ΔS°, and ΔG° for this spontaneous process?
ΔH° is negative, ΔS° is negative, and ΔG° is negative.
ΔH° is positive, ΔS° is positive, and ΔG° is positive.
ΔH° is positive, ΔS° is positive, and ΔG° is negative.
ΔH° is positive, ΔS° is negative, and ΔG° is positive.
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If the reaction quotient, Q, is less than the equilibrium constant, Keq, what is true about the Gibbs free energy?
ΔG° > 0 and the reaction is non-spontaneous in the forward direction.
ΔG° < 0 and the reaction is thermodynamically favored in the forward direction.
ΔG° = 0 and the system is at equilibrium.
ΔG° > 0 and the reaction is thermodynamically favored in the reverse direction.
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2SO2 (g) + O2 (g) ⇌ 2SO3 (g)
ΔH500 = -197 kJ mol-1
ΔG500 = -75 kJ mol-1
When equal volumes of sulfur dioxide and oxygen each at 1 atm, are mixed in a closed container at 500 K, no formation of SO3 (g) is observed.
Which of the following best explains the observation?
The reverse reaction has a lower activation energy than the forward reaction.
The reaction has a very small equilibrium constant at 500 K, so almost no product is formed.
A high activation energy makes the forward reaction extremely slow at 500 K.
The sulfur dioxide and oxygen must be mixed in a 2:1 ratio for the reaction to occur.
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A student observes that the equilibrium constant for a reaction is less than 1.0 at temperatures below 300 K but greater than 1.0 at temperatures above 300 K.
What can the student conclude about the values of ΔHo and ΔSo for the reaction?
(Assume that ΔHo and ΔSo are independent of temperature.)
ΔHo is positive and ΔSo is positive
ΔHo is positive and ΔSo is negative
ΔHo is negative and ΔSo is positive
ΔHo is negative and ΔSo is negative
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CaO (s) ⇌ Ca (s) + ½O2 (g) Keq = 4.0 x 10-10 at 1500 K
C (s) + ½O2 (g) ⇌ CO (g) Keq = 3.0 x 1027 at 1500 K
At 1500 K, the formation of Ca (s) and O₂ (g) from CaO (s) is not thermodynamically favorable. Adding C (s) at this temperature drives the reaction forward.
Based on the information provided, which of the following gives the value of Keq and the sign of ΔG° for the reaction below
CaO (s) + C (s) ⇌ Ca (s) + CO (g)
Keq | ΔG° | |
|---|---|---|
A | 1 x 10-35 | Positive |
B | 1 x 10-35 | Negative |
C | 1 x 1018 | Positive |
D | 1 x 1018 | Negative |
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