105 practice questions covering Equilibrium, organised into 0 topics.
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Sample questions
Q1
easy
Which of the following statements best describes the dynamic nature of chemical equilibrium?
AThe concentration of products remains constant while reactants are consumed.
BThe concentrations of reactants and products are equal at equilibrium.
CThe concentration of reactants is zero at equilibrium.
DThe rates of the forward and reverse reactions are equal at equilibrium.
Show answer and explanation
Correct answer: D - The rates of the forward and reverse reactions are equal at equilibrium.
At equilibrium, the rates of the forward and reverse reactions are equal, resulting in no net change in concentrations. Concentrations need not be equal, just constant.
Q2
medium
Three monobasic acids have these ionisation constants: HA, $1.8 \times 10^{-5}$; HB, $6.8 \times 10^{-4}$; HC, $4.9 \times 10^{-10}$. Arrange their conjugate bases in increasing order of basic strength.
A$\text{B}^- < \text{A}^- < \text{C}^-$
B$\text{C}^- < \text{A}^- < \text{B}^-$
C$\text{A}^- < \text{B}^- < \text{C}^-$
D$\text{B}^- < \text{C}^- < \text{A}^-$
Show answer and explanation
Correct answer: A - $\text{B}^- < \text{A}^- < \text{C}^-$
For a conjugate pair $K_a \times K_b = K_w$, so the larger the $K_a$ of the acid, the smaller the $K_b$ of its conjugate base. Acid strength runs HB, then HA, then HC, so base strength runs in the reverse order and increases as $\text{B}^-$, $\text{A}^-$, $\text{C}^-$. Ordering the bases the same way as the acids, or swapping the middle term, ignores the inverse relationship.
Q3
medium
Solid ammonium hydrogen sulphide decomposes in a closed vessel: $\text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g)$. Which expression for $K_p$ is correct?
Correct answer: B - $K_p = p_{\text{NH}_3} \times p_{\text{H}_2\text{S}}$
The equilibrium is heterogeneous, and a pure solid has a fixed concentration that is absorbed into the constant, so the ammonium hydrogen sulphide term does not appear. Both products are gases with a coefficient of one, so $K_p$ is simply the product of their partial pressures. Including the solid, averaging the two pressures, or writing a ratio of the two products all misuse the definition.
Q4
easy
Which of the following is true about buffer solutions?
AThey significantly change pH when a strong acid or base is added.
BThey do not follow the Henderson-Hasselbalch equation.
CThey can only be prepared using strong acids and strong bases.
DThey consist of a weak acid and its conjugate base or a weak base and its conjugate acid.
Show answer and explanation
Correct answer: D - They consist of a weak acid and its conjugate base or a weak base and its conjugate acid.
Buffer solutions consist of a weak acid and its conjugate base or a weak base and its conjugate acid, allowing them to resist changes in pH when small amounts of strong acids or bases are added. They follow the Henderson-Hasselbalch equation which is specific for calculating pH of buffer solutions.
Q5
easy
For $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$, a flask at equilibrium contains $[\text{N}_2\text{O}_4] = 0.50\ \text{M}$ and $[\text{NO}_2] = 0.20\ \text{M}$. Calculate $K_c$.
A$0.40$
B$12.5$
C$0.080$
D$0.16$
Show answer and explanation
Correct answer: C - $0.080$
$K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \frac{(0.20)^2}{0.50} = \frac{0.040}{0.50} = 0.080$. The value $0.40$ comes from omitting the square on the $\text{NO}_2$ term, $12.5$ from inverting the whole expression, and $0.16$ from squaring the entire ratio instead of only the $\text{NO}_2$ concentration.
Q6
medium
Consider the equilibrium reaction: 2SO2(g) + O2(g) ⇌ 2SO3(g). What happens when the volume of the container is halved at constant temperature?
AThe equilibrium shifts to the right, forming more SO3.
BThe equilibrium constant K increases.
CThe equilibrium remains unchanged.
DThe equilibrium shifts to the left, decomposing SO3 into SO2 and O2.
Show answer and explanation
Correct answer: A - The equilibrium shifts to the right, forming more SO3.
Decreasing the volume increases pressure, favoring the side with fewer moles of gas. Here, the right side has fewer gas moles, so the equilibrium shifts right.
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