4.9 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. Having to look up information or the solution should be an indicator that the previous layers (1–5) in the Structured Learning Approach have not been mastered.

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?
1. 0.2
2. 0.4
3. 0.8
4. 1.2
5. 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 K_p for the following reaction at 5 °C?

\[\mathrm{N_2O_4}(g) \rightleftharpoons \mathrm{2NO_2}(g) \quad K = 15\]
1. 15.0
2. 289.6
3. 6.2
4. 367.0
5. 342.4
Solution

Answer: E

Concept: K to K_p Relation

Is the reaction balanced?

Solve for K_p

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 K_p for the following reaction at 25 °C?

\[\mathrm{NO_2}(g) \rightleftharpoons \mathrm{N_2O_4}(g) \quad K = 215.98\]
1. 8.83
2. 12.23
3. 443.09
4. 5.28 × 10³
5. 5.35 × 10⁵
Solution

Answer: A

Concept: K to K_p Relation

Is the reaction balanced?

Solve for K_p

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?

\[\mathrm{A}(aq) + \mathrm{B}(aq) \rightleftharpoons \mathrm{2C}(aq)\]
1. 0.12
2. 0.25
3. 3.5
4. 6.8
5. 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

\[\begin{align*} K &= \dfrac{[\mathrm{C}]^2}{[\mathrm{A}][\mathrm{B}]} \\[1.5ex] &= \dfrac{(0.68~M)^2}{(0.21~M)(0.21~M)} \\[1.5ex] &= 10.485 \end{align*}\]
A reaction is found to be proceeding left towards equilibrium at a particular point in the reaction. Pick the true statement.
1. Q = K
2. Q > K
3. Q < K
4. rate_f = rate_r
5. 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?
1. 20
2. 33
3. 60
4. 80
5. 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?
1. The rates of the forward and reverse reactions are equal and remain constant
2. Concentrations of reactants and products are equal and remain constant
3. The rates of the forward and reverse reactions are not equal but remain constant
4. The forward and reverse reactions have stopped
5. none of these
Solution

Answer: A

Concept: Equilibrium
Pick the false statement regarding the following reaction at equilibrium.

\[\mathrm{H_2}(g) + \mathrm{I_2}(g) \rightleftharpoons \mathrm{HI}(g) \quad \Delta H > 0\]
1. Lowering the temperature drives the reaction right
2. Adding H₂ drives the reaction right
3. Removing HI drives the reaction right
4. For every 1 mol of H₂ consumed, 2 mol HI is produced
5. Decreasing P drives the reaction toward the right
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?

\[2\mathrm{PbS}(s) + \mathrm{3O_2}(g) \rightleftharpoons \mathrm{2PbO}(s) + 2\mathrm{SO_2}(g)\]
1. addition of heat
2. removal of heat
3. addition of PbS
4. removal of O₂
5. 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. –1.40 × 10^–3
2. 1.40 × 10^–3
3. 1.0
4. 714.3
5. 2.80 × 10³
Solution

Answer: D

Concept: Transforming K

When reversing a reaction, take the inverse of the equilibrium constant.

\[\begin{align*} K' &= \dfrac{1}{K} \\[1.5ex] &= \dfrac{1}{1.40\times 10^{-3}} \\[1.5ex] &= 714.28 \end{align*}\]

	H₂(g)	I₂(g)	2HI(g)
I	0.5	0.5	0
C	-x	-x	+2x
E	0.5-x	0.5-x	2x

	A(aq)	B(aq)	2C(aq)
I	0.55	0.55	0
C	-x	-x	+2x
E	0.55-x	0.21	2x