6.9 Practice Problems
Attempt these problems as if they were real exam questions in an exam environment.
Only look up information if you get severely stuck. Never look at the solution until you have exhausted all efforts to solve the problem.
If [H2] = 0.5 and [I2] = 0.5, initially, for the following reaction at 400 °C,
\[\mathrm{H_2}(g) + \mathrm{I_2}(g) \rightleftharpoons 2\mathrm{HI}(g) \quad K = 64.0\]
what is [HI] (in M) at equilibrium?
- 0.2
- 0.4
- 0.8
- 1.2
- 4.6
Solution
Answer: C
Concept: Solving Equilibrium Problems
Is the reaction balanced?
ICE Table
H2(g) + I2(g) ⇌ 2HI(g) I 0.5 0.5 0 C -x -x +2x E 0.5-x 0.5-x 2x Equilibrium Expression
\[\begin{align*} \dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} &= K \\[1.5ex] \dfrac{(2x)^2}{(0.5-x)(0.5-x)} &= 64.0 \\[1.5ex] \sqrt{\dfrac{(2x)^2}{(0.5-x)^2}} &= \sqrt{64.0} \\[1.5ex] \dfrac{2x}{0.5-x} &= 8 \\[1.5ex] 2x &= (8)(0.5-x) \\[1.5ex] 2x &= 4 - 8x \\[1.5ex] 10x &= 4 \\[1.5ex] x &= 0.4 \end{align*}\]
HI Concentration
\[\begin{align*} [\mathrm{HI}]_{\mathrm{eq}} &= 2x \\[1.5ex] &= 2(0.4~M) \\[1.5ex] &= 0.8~M \end{align*}\]
What is Kp for the following reaction at 5 °C?
\[\mathrm{N_2O_4}(g) \rightleftharpoons \mathrm{2NO_2}(g) \quad K = 15\]
- 15.0
- 289.6
- 6.2
- 367.0
- 342.4
Solution
Answer: E
Concept: K to Kp Relation
Is the reaction balanced?
Solve for Kp
Note that Δn is 1.
\[\begin{align*} K_P &= K (RT)^{\Delta n} \\[1.5ex] &= 15 (0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}} \times 278.15~\mathrm{K})^{1}\\[1.5ex] &= 366 \end{align*}\]
What is Kp for the following reaction at 25 °C?
\[\mathrm{NO_2}(g) \rightleftharpoons \mathrm{N_2O_4}(g) \quad K = 215.98\]
- 8.83
- 12.23
- 443.09
- 5.28 × 103
- 5.35 × 105
Solution
Answer: A
Concept: K to Kp Relation
Is the reaction balanced?
Solve for Kp
First, balance the equation and then note that Δn is –1.
\[\begin{align*} K_P &= K (RT)^{\Delta n} \\[1.5ex] &= 215.98 (0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}} \times 278.15~\mathrm{K})^{-1}\\[1.5ex] &= 8.827 \end{align*}\]
A solution initially contains 0.55 M of A and B. The following balanced reaction occurs and the equilibrium concentration for B is found to be 0.21 M. What is K for this reaction?
\[\mathrm{A}(aq) + \mathrm{B}(aq) \rightleftharpoons \mathrm{2C}(aq)\]
- 0.12
- 0.25
- 3.5
- 6.8
- 10.5
Solution
Answer: E
Concept: Solving Equilibrium Problems
Is the reaction balanced?
ICE Table
A(aq) + B(aq) ⇌ 2C(aq) I 0.55 0.55 0 C -x -x +2x E 0.55-x 0.21 2x Determine x
We are given the initial and equilibrium concentration of B. We can directly determine x.
\[\begin{align*} 0.55 - x &= 0.21 \\[1.5ex] x &= 0.34 \end{align*}\]
Find Equilibrium Concentrations
\[\begin{align*} [\mathrm{A}]_{\mathrm{eq}} &= 0.55~M - x \\[1.5ex] &= 0.55~M - 0.34~M \\[1.5ex] &= 0.21~M \\[2.5ex] [\mathrm{B}]_{\mathrm{eq}} &= 0.21~M \\[2.5ex] [\mathrm{C}]_{\mathrm{eq}} &= 2x \\[1.5ex] &= 2(0.34~M) \\[1.5ex] &= 0.68~M \end{align*}\]
Find K
\[\begin{align*} K &= \dfrac{[\mathrm{C}]^2}{[\mathrm{A}][\mathrm{B}]} \\[1.5ex] &= \dfrac{(0.68~M)^2}{(0.21~M)(0.21~M)} \\[1.5ex] &= 10.485 \end{align*}\]
- 0.12
A reaction is found to be proceeding left towards equilibrium at a particular point in the reaction. Pick the true statement.
- Q = K
- Q > K
- Q < K
- ratef = rater
- none of these
Solution
Answer: B
Concept: Q vs K
The question states that a reaction is proceeding left Therefore, Q > K at that point in time.
The following reaction is carried out in a 1.0 L vessel:
\[\mathrm{SO_2}(g) + \mathrm{H_2O}(g) \rightleftharpoons \mathrm{H_2SO_3}(g) \quad K = 0.115\]
What is the mass (in g) of the product at equilibrium if there is 40.0 g of SO2 (m.m. = 64.066 g mol–1) and 100.0 g of H2O at equilibrium?
- 20
- 33
- 60
- 80
- 100
Solution
Answer: B
Concept: Q vs K
Is the reaction balanced?
We begin with the equilibrium amounts of the reactants SO2 and H2O. We are also provided with the equilibrium constant. Therefore, we should be able to directly solve for the H2SO3 product concentration at equilibrium.
Find Moles
\[\begin{align*} n_{\mathrm{SO_2}} &= 40.0~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{64.066~\mathrm{g}} \right ) \\[1.5ex] &= 0.62436~\mathrm{mol} \\[2.5ex] n_{\mathrm{H_2O}} &= 100.0~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) \\[1.5ex] &= 5.54939~\mathrm{mol} \end{align*}\]
Find Molar Concentrations
Convert the given masses into molar concentrations. We are dealing with a gas phase reaction and gases fill the container they are in. Treating the gases as ideal, we can use the volume of the flask (here, 1 L) for the volume of each gas.
\[\begin{align*} [\mathrm{SO_2}]_{\mathrm{eq}} &= \dfrac{n}{V} \\[1.5ex] &= \dfrac{0.62436~\mathrm{mol}}{\mathrm{1~\mathrm{L}}}\\[1.5ex] &= 0.62436~M \\[2.5ex] [\mathrm{H_2O}]_{\mathrm{eq}} &= \dfrac{n}{V} \\[1.5ex] &= \dfrac{5.54939~\mathrm{mol}}{\mathrm{1~\mathrm{L}}}\\[1.5ex] &= 5.54939~M \end{align*}\]
Find Molar Concentration of Product
\[\begin{align*} \dfrac{[\mathrm{H_2SO_3}]}{[\mathrm{SO_2}][\mathrm{H_2O}]} &= K \\[1.5ex] \dfrac{x}{(0.62436~M)(5.54939~M)} &= 0.115 \\[1.5ex] x &= 0.39845~M \longrightarrow \\[1.5ex] [\mathrm{H_2SO_3}]_{\mathrm{eq}} &= 0.39845~M \end{align*}\]
\[\begin{align*} m_{\mathrm{H_2SO_3}} &= \left ( \dfrac{0.39845~\mathrm{mol}}{\mathrm{L}}\right ) \left ( \dfrac{82.07~\mathrm{g}}{\mathrm{mol}}\right ) \\ &= 32.67 \end{align*}\]
Which statement is true regarding chemical equilibrium?
- The rates of the forward and reverse reactions are equal and remain constant
- Concentrations of reactants and products are equal and remain constant
- The rates of the forward and reverse reactions are not equal but remain constant
- The forward and reverse reactions have stopped
- none of these
Solution
Answer: A
Concept: Equilibrium
Pick the false statement regarding the following reaction at equilibrium.
\[\mathrm{H_2}(g) + \mathrm{I_2}(g) \rightleftharpoons 2\mathrm{HI}(g) \quad \Delta H > 0\]
- Lowering the temperature drives the reaction right
- Adding H2 drives the reaction right
- Removing HI drives the reaction right
- For every 1 mol of H2 consumed, 2 mol HI is produced
- Decreasing P has no effect on the reaction direction
Solution
Answer: A
Concept: Le Chatelier’s Principle
The given reaction is endothermic since ΔH > 0 (it absorbs heat). Therefore, we can treat heat as a reactant. If heat is taken away, the reaction would shift to the left. Statement A is false
Which action will drive the following endothermic reaction toward the right?
\[2\mathrm{PbS}(s) + \mathrm{3O_2}(g) \rightleftharpoons \mathrm{2PbO}(s) + 2\mathrm{SO_2}(g)\]
- addition of heat
- removal of heat
- addition of PbS
- removal of O2
- none of these
Solution
Answer: A
Concept: Le Chatelier’s Principle
The given reaction is endothermic since ΔH > 0 (it absorbs heat). Therefore, we can treat heat as a reactant. If heat is added, it will drive the reaction to the right.
What is the equilibrium constant for the reverse of the following reaction?
\[\mathrm{A}+\mathrm{2B} \longrightarrow \mathrm{3C} \qquad K = 1.40 \times 10^{-3}\]
- –1.40 × 10–3
- 1.40 × 10–3
- 1.0
- 714.3
- 2.80 × 103
Solution
Answer: D
Concept: Transforming K
When reversing a reaction, take the inverse of the equilibrium constant.
\[\begin{align*} K' &= \dfrac{1}{K} \\[1.5ex] &= \dfrac{1}{1.40\times 10^{-3}} \\[1.5ex] &= 714.28 \end{align*}\]