Thermodynamics — Hard Practice Quiz
A Chemistry cheat sheet for Thermodynamics — every key formula with its symbols defined — plus a hard-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
Which of the following expressions correctly relates the Gibbs free energy change of a system, $\Delta G$, to the total entropy change of the universe, $\Delta S_{\text{univ}}$, for a process occurring at constant temperature $T$ and pressure $P$?
- $\Delta G = -T\Delta S_{\text{univ}}$
- $\Delta G = T\Delta S_{\text{univ}}$
- $\Delta G = -\frac{\Delta S_{\text{univ}}}{T}$
- $\Delta G = \Delta H_{\text{sys}} + T\Delta S_{\text{univ}}$
Answer: $\Delta G = -T\Delta S_{\text{univ}}$
Consider a reaction where the standard enthalpy change is $\Delta H^\circ = -150 \text{ kJ}$ and the standard entropy change is $\Delta S^\circ = -50 \text{ J/K}$. Calculate the standard Gibbs free energy change, $\Delta G^\circ$, for this reaction at $298 \text{ K}$. Is the reaction spontaneous under standard conditions at this temperature?
- $\Delta G^\circ = -135.1 \text{ kJ}$; Spontaneous
- $\Delta G^\circ = -164.9 \text{ kJ}$; Spontaneous
- $\Delta G^\circ = 135.1 \text{ kJ}$; Non-spontaneous
- $\Delta G^\circ = 164.9 \text{ kJ}$; Non-spontaneous
Answer: $\Delta G^\circ = -135.1 \text{ kJ}$; Spontaneous
For a certain reaction, the equilibrium constant $K$ is $1.0 \times 10^5$ at $298 \text{ K}$. If the standard enthalpy change for the reaction is $\Delta H^\circ = -75 \text{ kJ/mol}$, what is the approximate equilibrium constant $K'$ at $373 \text{ K}$? Assume $\Delta H^\circ$ and $\Delta S^\circ$ are constant over this temperature range. (Use $R = 8.314 \text{ J/(mol} \cdot \text{K)}$)
- $4.4 \times 10^7$
- $2.3 \times 10^3$
- $1.0 \times 10^5$
- $1.8 \times 10^6$
Answer: $4.4 \times 10^7$
A $1.0 \text{ mol}$ sample of supercooled water at $-10 \text{ }^\circ\text{C}$ freezes irreversibly to ice at $-10 \text{ }^\circ\text{C}$ and $1 \text{ atm}$. Given that the enthalpy of fusion for water at $0 \text{ }^\circ\text{C}$ is $\Delta H_{\text{fus}} = 6.01 \text{ kJ/mol}$, the molar heat capacity of liquid water is $C_{p,l} = 75.3 \text{ J/(mol} \cdot \text{K)}$, and for ice is $C_{p,s} = 37.6 \text{ J/(mol} \cdot \text{K)}$. Calculate the total entropy change of the universe, $\Delta S_{\text{univ}}$, for this process. (Hint: Consider a reversible path from supercooled water to ice at $-10 \text{ }^\circ\text{C}$)
- $0.80 \text{ J/K}$
- $-0.80 \text{ J/K}$
- $2.79 \text{ J/K}$
- $-22.00 \text{ J/K}$
Answer: $0.80 \text{ J/K}$
A system undergoes a reversible isothermal process at temperature $T$, absorbing $q_{\text{rev}}$ amount of heat. If the initial number of microstates is $W_1$, what is the final number of microstates, $W_2$, in terms of $q_{\text{rev}}$, $T$, and Boltzmann's constant $k$?
- $W_2 = W_1 e^{\frac{q_{\text{rev}}}{kT}}$
- $W_2 = W_1 e^{\frac{kT}{q_{\text{rev}}}}$
- $W_2 = W_1 + \frac{q_{\text{rev}}}{kT}$
- $W_2 = \frac{q_{\text{rev}}}{kT} \ln W_1$
Answer: $W_2 = W_1 e^{\frac{q_{\text{rev}}}{kT}}$
For the reaction $2\text{NO(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}$, given the standard Gibbs free energies of formation at $298 \text{ K}$: $\Delta G^\circ_f(\text{NO(g)}) = 86.55 \text{ kJ/mol}$ and $\Delta G^\circ_f(\text{NO}_2\text{(g)}) = 51.31 \text{ kJ/mol}$. Calculate the equilibrium constant $K$ for this reaction at $298 \text{ K}$. (Use $R = 8.314 \text{ J/(mol} \cdot \text{K)}$)
- $2.3 \times 10^{12}$
- $4.4 \times 10^{-13}$
- $1.0 \times 10^{-7}$
- $7.0 \times 10^1$
Answer: $2.3 \times 10^{12}$
A chemical reaction has a standard enthalpy change of $\Delta H^\circ = -250 \text{ kJ}$ and a standard entropy change of $\Delta S^\circ = -100 \text{ J/K}$. What is the maximum non-PV work that can be extracted from this reaction at $350 \text{ K}$?
- $215 \text{ kJ}$
- $-215 \text{ kJ}$
- $285 \text{ kJ}$
- $-285 \text{ kJ}$
Answer: $215 \text{ kJ}$
For a reaction with a positive standard enthalpy change ($\Delta H^\circ > 0$) and a positive standard entropy change ($\Delta S^\circ > 0$), which of the following statements about its spontaneity is true?
- The reaction is spontaneous at high temperatures.
- The reaction is spontaneous at low temperatures.
- The reaction is spontaneous at all temperatures.
- The reaction is non-spontaneous at all temperatures.
Answer: The reaction is spontaneous at high temperatures.
For the reaction $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{SO}_3\text{(g)}$, given the standard molar entropies at $298 \text{ K}$: $S^\circ(\text{SO}_2\text{(g)}) = 248.2 \text{ J/(mol} \cdot \text{K)}$, $S^\circ(\text{O}_2\text{(g)}) = 205.1 \text{ J/(mol} \cdot \text{K)}$, and $S^\circ(\text{SO}_3\text{(g)}) = 256.8 \text{ J/(mol} \cdot \text{K)}$. If the standard enthalpy change for this reaction is $\Delta H^\circ = -197.8 \text{ kJ}$, at what temperature range is this reaction spontaneous under standard conditions?
- $T < 1053 \text{ K}$
- $T > 1053 \text{ K}$
- Spontaneous at all temperatures.
- Non-spontaneous at all temperatures.
Answer: $T < 1053 \text{ K}$
A reaction has a standard Gibbs free energy change of $\Delta G^\circ = -30.0 \text{ kJ/mol}$ at $298 \text{ K}$. If the reaction quotient $Q$ is $0.01$ at this temperature, what is the maximum non-PV work that can be obtained from the reaction under these nonstandard conditions? (Use $R = 8.314 \text{ J/(mol} \cdot \text{K)}$)
- $41.4 \text{ kJ/mol}$
- $30.0 \text{ kJ/mol}$
- $18.6 \text{ kJ/mol}$
- $-41.4 \text{ kJ/mol}$
Answer: $41.4 \text{ kJ/mol}$
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