Reaction Equilibrium — Hard Practice Quiz
A Chemistry cheat sheet for Reaction Equilibrium — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.
Formulas & key concepts
Equilibrium Constant (Concentration): Ratio of product concentrations to reactant concentrations at equilibrium, raised to stoichiometric coefficients.
Where: \(K_c\) = equilibrium constant
Haber Process Equilibrium: Specific equilibrium constant expression for \(N_2 + 3H_2 \rightleftharpoons 2NH_3\).
Equilibrium Constant (Pressure): Ratio of product partial pressures to reactant partial pressures at equilibrium.
Where: \(K_p\) = equilibrium constant (pressure)
Pressure-Concentration Relation: Relates partial pressure of a gas to its molar concentration.
Where: \(P\) = pressure, \([A]\) = concentration, \(R\) = gas constant, \(T\) = temperature
Kp-Kc Relationship: Relates equilibrium constants based on pressure and concentration.
Where: \(\Delta n\) = change in moles of gas
Change in Moles: Difference between sum of gaseous product coefficients and gaseous reactant coefficients.
Reaction Quotient: Same expression as \(K_c\) but calculated with current concentrations, not necessarily at equilibrium.
Where: \(Q_c\) = reaction quotient
Endothermic Reaction: Heat acts as a reactant; increasing temperature shifts equilibrium to products (right).
Exothermic Reaction: Heat acts as a product; increasing temperature shifts equilibrium to reactants (left).
Practice quiz
For the Haber process, $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, if the equilibrium constant $K_c$ is $0.50 \text{ M}^{-2}$ at $400 \text{ K}$, what is the value of $K_p$ at the same temperature? Use $R = 0.0821 \text{ L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$.
- $0.00046$
- $16.42$
- $0.50$
- $0.000014$
Answer: $0.00046$
Consider the reaction $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$. If the volume of the reaction vessel is suddenly halved at constant temperature, what is the immediate effect on the values of $K_c$ and $K_p$?
- Both $K_c$ and $K_p$ will increase.
- Both $K_c$ and $K_p$ will decrease.
- $K_c$ will increase, but $K_p$ will remain unchanged.
- Both $K_c$ and $K_p$ will remain unchanged.
Answer: Both $K_c$ and $K_p$ will remain unchanged.
For the reaction $A(g) + B(g) \rightleftharpoons 2C(g)$, the equilibrium constant $K_c$ is $4.0$ at a specific temperature. If the initial concentrations are $[A] = 0.1 \text{ M}$, $[B] = 0.1 \text{ M}$, and $[C] = 0.3 \text{ M}$, in which direction will the reaction proceed to reach equilibrium?
- The reaction will proceed to the right (towards products).
- The reaction will proceed to the left (towards reactants).
- The reaction is already at equilibrium.
- The direction cannot be determined without knowing the temperature.
Answer: The reaction will proceed to the left (towards reactants).
For the reaction $2X(g) \rightleftharpoons Y(g) + Z(g)$, the equilibrium constant $K_p$ is $0.25$ at $300 \text{ K}$. What is the value of $K_c$ at this temperature? Use $R = 0.0821 \text{ L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$.
- $K_c = 0.25$
- $K_c = 0.25 \times (0.0821 \times 300)$
- $K_c = \frac{0.25}{(0.0821 \times 300)^2}$
- $K_c = \frac{0.25}{(0.0821 \times 300)}$
Answer: $K_c = 0.25$
Consider the exothermic reaction $2NO_2(g) \rightleftharpoons N_2O_4(g) + \text{Heat}$. If the temperature of the system is increased, what will be the effect on the equilibrium constant $K_c$ and the position of equilibrium?
- $K_c$ will increase, and the equilibrium will shift to the right.
- $K_c$ will decrease, and the equilibrium will shift to the left.
- $K_c$ will remain unchanged, but the equilibrium will shift to the left.
- $K_c$ will decrease, and the equilibrium will shift to the right.
Answer: $K_c$ will decrease, and the equilibrium will shift to the left.
For the gaseous reaction $A(g) \rightleftharpoons 2B(g)$, derive the relationship between $K_p$ and $K_c$ using the ideal gas law relation $P_i = [i]RT$.
- $K_p = K_c(RT)^{-1}$
- $K_p = K_c(RT)$
- $K_p = K_c(RT)^2$
- $K_p = K_c$
Answer: $K_p = K_c(RT)$
For a reversible reaction at $298 \text{ K}$, the equilibrium constant $K_p$ is $1.0 \times 10^{-3}$ and $K_c$ is $4.0 \times 10^{-5}$. What is the value of $\Delta n$ (change in moles of gas) for this reaction? Use $R = 0.0821 \text{ L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$.
- $\Delta n = -1$
- $\Delta n = 0$
- $\Delta n = 1$
- $\Delta n = 2$
Answer: $\Delta n = 1$
Consider the reaction $A(g) + B(g) \rightleftharpoons C(g)$. What is the effect of adding an inert gas, such as Argon, to this system at constant temperature?
- Adding inert gas at constant volume shifts the equilibrium to the right, while adding it at constant pressure has no effect.
- Adding inert gas at constant volume has no effect, while adding it at constant pressure shifts the equilibrium to the left.
- Adding inert gas at constant volume shifts the equilibrium to the left, while adding it at constant pressure shifts it to the right.
- Adding inert gas at constant volume has no effect, and adding it at constant pressure also has no effect.
Answer: Adding inert gas at constant volume has no effect, while adding it at constant pressure shifts the equilibrium to the left.
For the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$, the equilibrium constant $K_c$ is $64$ at $450 \text{ K}$. If $0.10 \text{ mol}$ of $H_2$ and $0.10 \text{ mol}$ of $I_2$ are initially placed in a $1.0 \text{ L}$ vessel, what is the equilibrium concentration of $HI$?
- $0.02 \text{ M}$
- $0.08 \text{ M}$
- $0.16 \text{ M}$
- $0.20 \text{ M}$
Answer: $0.16 \text{ M}$
Consider an endothermic reaction $A(g) \rightleftharpoons B(g) + C(g)$. If the temperature of the system is increased and the total pressure is simultaneously decreased, what will be the overall effect on the equilibrium position and the value of $K_c$?
- Equilibrium shifts to the left, and $K_c$ decreases.
- Equilibrium shifts to the right, and $K_c$ increases.
- Equilibrium shifts to the right, but $K_c$ remains unchanged.
- Equilibrium shifts to the left, and $K_c$ increases.
Answer: Equilibrium shifts to the right, and $K_c$ increases.
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