Thermodynamics — Practice Quiz
A Chemistry cheat sheet for Thermodynamics — every key formula with its symbols defined — plus a medium-level practice quiz to test recall.
Formulas & key concepts
Entropy Change: Relates entropy change to heat absorbed/released in a reversible process at constant T.
Where: \(q_{\text{rev}}\) = reversible heat, \(T\) = temperature
Second Law of Thermodynamics: For irreversible process \(\Delta S_{\text{univ}} > 0\), for reversible process \(\Delta S_{\text{univ}} = 0\).
Boltzmann Entropy: Relates entropy to number of microstates.
Where: \(k\) = Boltzmann constant, \(W\) = number of microstates
Standard Entropy Change: Calculate from standard molar entropies.
Entropy Change of Surroundings: At constant T and P.
Gibbs Free Energy: Fundamental equation at constant T.
Where: \(\Delta G\) = free energy change
Standard Free Energy Change: Calculate from standard free energies of formation.
Free Energy and Reversibility: For reversible process \(\Delta G = 0\), for irreversible process \(\Delta G < 0\).
Free Energy and Work: Maximum work a process can perform.
Where: \(w_{\text{max}}\) = maximum work
Free Energy Under Nonstandard Conditions: Relates \(\Delta G\) to reaction quotient.
Where: \(Q\) = reaction quotient
Free Energy and Equilibrium Constant: Relates standard free energy to equilibrium constant.
Where: \(K\) = equilibrium constant
Practice quiz
A system absorbs $100 \text{ J}$ of heat reversibly at a constant temperature of $300 \text{ K}$. What is the entropy change of the system?
- $0.333 \text{ J/K}$
- $-0.333 \text{ J/K}$
- $3.00 \text{ J/K}$
- $-3.00 \text{ J/K}$
Answer: $0.333 \text{ J/K}$
For a certain process, the entropy change of the system is $50 \text{ J/K}$ and the entropy change of the surroundings is $-60 \text{ J/K}$. Which of the following statements is true regarding this process?
- The process is spontaneous.
- The process is non-spontaneous.
- The process is at equilibrium.
- The process is reversible.
Answer: The process is non-spontaneous.
If a system has $W = 1$ microstate, what is its entropy according to Boltzmann's equation?
- $S = k$
- $S = 0$
- $S = \ln k$
- $S = \text{undefined}$
Answer: $S = 0$
An exothermic reaction releases $250 \text{ kJ}$ of heat to the surroundings at a constant temperature of $298 \text{ K}$. What is the entropy change of the surroundings?
- $839 \text{ J/K}$
- $-839 \text{ J/K}$
- $0.839 \text{ J/K}$
- $-0.839 \text{ J/K}$
Answer: $839 \text{ J/K}$
Which of the following conditions guarantees a spontaneous reaction at all temperatures?
- $\Delta H > 0$ and $\Delta S > 0$
- $\Delta H < 0$ and $\Delta S < 0$
- $\Delta H < 0$ and $\Delta S > 0$
- $\Delta H > 0$ and $\Delta S < 0$
Answer: $\Delta H < 0$ and $\Delta S > 0$
A reaction has $\Delta H = -120 \text{ kJ/mol}$ and $\Delta S = -250 \text{ J/(mol} \cdot \text{K)}$. At what temperature range will this reaction be spontaneous?
- Spontaneous at all temperatures.
- Spontaneous at temperatures below $480 \text{ K}$.
- Spontaneous at temperatures above $480 \text{ K}$.
- Non-spontaneous at all temperatures.
Answer: Spontaneous at temperatures below $480 \text{ K}$.
If a chemical process has a Gibbs free energy change of $\Delta G = -50 \text{ kJ/mol}$, what is the maximum amount of non-PV work that can be extracted from this process?
- $50 \text{ kJ/mol}$
- $-50 \text{ kJ/mol}$
- $0 \text{ kJ/mol}$
- $100 \text{ kJ/mol}$
Answer: $50 \text{ kJ/mol}$
For a reaction at $298 \text{ K}$ with $\Delta G^\circ = -30 \text{ kJ/mol}$ and a reaction quotient $Q = 0.1$, what is the value of $\Delta G$? (Assume $R = 8.314 \text{ J/(mol} \cdot \text{K)}$)
- Approximately $-35.7 \text{ kJ/mol}$
- Approximately $-24.3 \text{ kJ/mol}$
- Approximately $-30.0 \text{ kJ/mol}$
- Approximately $35.7 \text{ kJ/mol}$
Answer: Approximately $-35.7 \text{ kJ/mol}$
If the standard Gibbs free energy change for a reaction is $\Delta G^\circ = 0$, what can be said about its equilibrium constant $K$ at standard conditions?
- $K = 0$
- $K = 1$
- $K > 1$
- $K < 1$
Answer: $K = 1$
Consider the reaction $2A(g) + B(g) \rightarrow C(g)$. Given the standard molar entropies: $S^\circ(A) = 150 \text{ J/(mol} \cdot \text{K)}$, $S^\circ(B) = 200 \text{ J/(mol} \cdot \text{K)}$, $S^\circ(C) = 300 \text{ J/(mol} \cdot \text{K)}$. Calculate the standard entropy change for the reaction, $\Delta S^\circ$.
- $-200 \text{ J/(mol} \cdot \text{K)}$
- $200 \text{ J/(mol} \cdot \text{K)}$
- $-50 \text{ J/(mol} \cdot \text{K)}$
- $50 \text{ J/(mol} \cdot \text{K)}$
Answer: $-200 \text{ J/(mol} \cdot \text{K)}$
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