Unit 5 of 5
Study guide for CLEP CLEP Chemistry — Unit 5: Thermodynamics and Kinetics. Practice questions, key concepts, and exam tips.
51
Practice Questions
49
Flashcards
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Key Topics
Try these 5 questions from this unit. Sign up for full access to all 51.
What is the standard reduction potential of the half-reaction Cu2+ + 2e- → Cu?
-0.34 V
-0.74 V
+0.34 V
+0.74 V
+0.50 V
Answer: C — +0.34 V is correct because the standard reduction potential of the half-reaction Cu2+ + 2e- → Cu is +0.34 V. The most tempting distractor, -0.34 V, would be the oxidation potential.
What is the effect of increasing temperature on the value of Kc for an exothermic reaction?
Increase
Decrease then increase
No effect
Increase then decrease
Decrease
Answer: E — Increasing temperature decreases the value of Kc for an exothermic reaction, as the equilibrium shifts to the left to consume heat and re-establish equilibrium. This occurs because exothermic reactions release heat, so higher temperatures favor the reverse reaction, reducing Kc.
Consider a system where ΔH = 50 kJ and TΔS = 70 kJ at 300 K. What is the sign of ΔG?
Positive
Zero
Cannot be determined
Negative
Depends on the specific reaction
Answer: D — Negative is correct because ΔG = ΔH - TΔS = 50 kJ - 70 kJ = -20 kJ, which is negative. A positive ΔG is incorrect as it would indicate non-spontaneity.
What is the main assumption of Hess's law?
The reaction is reversible
The reaction is irreversible
Enthalpy is a state function
The reaction is exothermic
The reaction is endothermic
Answer: C — Enthalpy is a state function is correct because Hess's law relies on the fact that enthalpy is a state function, meaning its value depends only on the initial and final states of the system, not on the path taken.
Which of the following equations describes the first law of thermodynamics?
ΔE = q + w
ΔE = q - w
ΔG = ΔH - TΔS
ΔH = ΔE + pΔV
ΔS = q / T
Answer: A — ΔE = q + w is correct because ΔE equals the sum of heat and work..
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