Equilibrium Chapter-Wise Test 15

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The solubility of \( \ce{PbI2} \) in 0.05 M \( \ce{KI} \) is \( 1.0 \times 10^{-4} \, \text{M} \). What is its \( K_{sp} \)?

For \( \ce{PbI2 <=> Pb^{2+} + 2I-} \), \( [\ce{Pb^{2+}}] = 1.0 \times 10^{-4} \), \( [\ce{I-}] = 0.05 + 2 \times 1.0 \times 10^{-4} \approx 0.05 \, \text{M} \). \( K_{sp} = [\ce{Pb^{2+}}][\ce{I-}]^2 = (1.0 \times 10^{-4}) (0.05)^2 = 2.5 \times 10^{-7} \).

\( 1.0 \times 10^{-4} \)
\( 5.0 \times 10^{-6} \)
\( 2.5 \times 10^{-7} \)
\( 4.0 \times 10^{-8} \)
3

For \( \ce{2NO(g) <=> N2(g) + O2(g)} \), \( K_c = 0.01 \) at 300 K. If 0.4 mol \( \ce{NO} \) and 0.1 mol \( \ce{N2} \) are in a 2 L vessel, what is \( [\ce{O2}] \) at equilibrium?

Initial: \( [\ce{NO}] = \frac{0.4}{2} = 0.2 \, \text{M} \), \( [\ce{N2}] = \frac{0.1}{2} = 0.05 \, \text{M} \), \( [\ce{O2}] = 0 \). Let \( x = [\ce{O2}] \), \( [\ce{NO}] = 0.2 - 2x \), \( [\ce{N2}] = 0.05 + x \). \( K_c = \frac{[\ce{N2}][\ce{O2}]}{[\ce{NO}]^2} = \frac{(0.05 + x)x}{(0.2 - 2x)^2} = 0.01 \), \( x \approx 0.004 \, \text{M} \).

0.004 M
0.01 M
0.02 M
0.05 M
1

For \( \ce{A(g) + 2B(g) <=> 2C(g)} \), \( K_p = 0.25 \) at 400 K. If the total pressure at equilibrium is 4 atm and \( P_{\ce{A}} = 1 \, \text{atm} \), what is \( P_{\ce{C}} \)?

Total pressure = \( P_{\ce{A}} + P_{\ce{B}} + P_{\ce{C}} = 4 \), \( P_{\ce{B}} + P_{\ce{C}} = 3 \). \( K_p = \frac{(P_{\ce{C}})^2}{P_{\ce{A}} (P_{\ce{B}})^2} = \frac{(P_{\ce{C}})^2}{1 \cdot (3 - P_{\ce{C}})^2} = 0.25 \), \( P_{\ce{C}} = 0.5 (3 - P_{\ce{C}}) \), \( P_{\ce{C}} = 1.5 - 0.5 P_{\ce{C}} \), \( 1.5 P_{\ce{C}} = 1.5 \), \( P_{\ce{C}} = 1 \, \text{atm} \).

2.0 atm
0.5 atm
1.0 atm
1.5 atm
3

In \( \ce{A(g) + 2B(g) <=> 2C(g)} \), \( K_c = 2 \) and \( [\ce{A}] = 0.2 \, \text{M} \), \( [\ce{B}] = 0.4 \, \text{M} \), \( [\ce{C}] = 0.4 \, \text{M} \) at equilibrium. What happens if \( \ce{C} \) is added?

Adding \( \ce{C} \) increases \( Q \), shifts left to reduce \( \ce{C} \).

Shifts right
No shift
\( K_c \) increases
Shifts left
4

In \( \ce{2A(g) + B(g) <=> 2C(g)} \), if \( K_c = 4 \) and the equilibrium mixture contains 0.2 mol \( \ce{A} \), 0.1 mol \( \ce{B} \), and 0.4 mol \( \ce{C} \) in a 1 L vessel, what happens if 0.1 mol \( \ce{A} \) is added?

Initial equilibrium: \( [\ce{A}] = 0.2 \, \text{M} \), \( [\ce{B}] = 0.1 \, \text{M} \), \( [\ce{C}] = 0.4 \, \text{M} \), \( K_c = \frac{(0.4)^2}{(0.2)^2(0.1)} = 4 \). After adding 0.1 mol \( \ce{A} \), \( [\ce{A}] = 0.3 \, \text{M} \), reaction shifts right to restore equilibrium.

Shifts left
No shift
\( K_c \) increases
Shifts right
4

The \( K_a \) of \( \ce{HF} \) is \( 6.8 \times 10^{-4} \). What is the \( K_b \) of \( \ce{F-} \) at 298 K?

\( K_w = K_a \times K_b = 1.0 \times 10^{-14} \), \( K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}} \approx 1.47 \times 10^{-11} \).

\( 6.8 \times 10^{-4} \)
\( 1.0 \times 10^{-14} \)
\( 3.4 \times 10^{-10} \)
\( 1.47 \times 10^{-11} \)
4

For the equilibrium \( \ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)} \), if the initial concentrations of \( \ce{CO} \) and \( \ce{H2O} \) are 0.2 M each and at equilibrium \( [\ce{CO2}] = 0.08 \, \text{M} \), what is \( K_c \)?

At equilibrium, \( [\ce{CO}] = [\ce{H2O}] = 0.2 - 0.08 = 0.12 \, \text{M} \), \( [\ce{CO2}] = [\ce{H2}] = 0.08 \, \text{M} \). Thus, \( K_c = \frac{[\ce{CO2}][\ce{H2}]}{[\ce{CO}][\ce{H2O}]} = \frac{(0.08)^2}{(0.12)^2} = \frac{0.0064}{0.0144} \approx 0.444 \).

0.8
0.444
2.25
1.0
2

A weak acid \( \ce{HX} \) (\( K_a = 1.0 \times 10^{-4} \)) has a 0.1 M solution mixed with 0.05 M \( \ce{NaX} \) in a 1:2 volume ratio. What is the pH?

Total volume = 3V, \( [\ce{HX}] = \frac{0.1V}{3V} = 0.0333 \, \text{M} \), \( [\ce{X-}] = \frac{0.05 \times 2V}{3V} = 0.0333 \, \text{M} \), \( \text{pH} = 4 + \log \frac{0.0333}{0.0333} = 4 \).

3.5
4.5
3.0
4.0
4

A weak acid \( \ce{HX} \) (\( K_a = 4.0 \times 10^{-6} \)) is mixed with 0.03 M \( \ce{NaX} \) in a 2:1 volume ratio (acid:salt). If \( [\ce{HX}] = 0.08 \, \text{M} \) after mixing, what is the pH?

Total volume = 3V, \( [\ce{X-}] = \frac{0.03V}{3V} = 0.01 \, \text{M} \), \( \text{p}K_a = 5.4 \), \( \text{pH} = 5.4 + \log \frac{0.01}{0.08} = 5.4 - 0.903 = 4.5 \).

5.4
4.0
4.5
5.0
3

For \( \ce{CO(g) + Cl2(g) <=> COCl2(g)} \), \( K_c = 25 \) at 500 K. If 0.2 mol \( \ce{CO} \) and 0.3 mol \( \ce{Cl2} \) are in a 1 L vessel, what is \( [\ce{COCl2}] \) at equilibrium?

Initial: \( [\ce{CO}] = 0.2 \, \text{M} \), \( [\ce{Cl2}] = 0.3 \, \text{M} \), \( [\ce{COCl2}] = 0 \). Let \( x = [\ce{COCl2}] \), \( [\ce{CO}] = 0.2 - x \), \( [\ce{Cl2}] = 0.3 - x \). \( K_c = \frac{[\ce{COCl2}]}{[\ce{CO}][\ce{Cl2}]} = \frac{x}{(0.2 - x)(0.3 - x)} = 25 \), \( x \approx 0.18 \, \text{M} \).

0.2 M
0.18 M
0.25 M
0.15 M
2

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