Equilibrium Chapter-Wise Test 16

Correct answer Carries: 4.

Wrong Answer Carries: -1.

For \( \ce{2NO2(g) <=> N2O4(g)} \), if \( K_p = 4.0 \) at 350 K (\( R = 0.0831 \, \text{bar L/mol K} \)), what is \( K_c \)?

\( K_p = K_c (RT)^{\Delta n} \), \( \Delta n = 1 - 2 = -1 \), \( RT = 0.0831 \times 350 \approx 29.085 \). Thus, \( K_c = K_p \times RT = 4.0 \times 29.085 \approx 116.34 \).

116.34
4.0
0.137
29.085
1

For \( \ce{2A(g) + B(g) <=> 2C(g)} \), \( K_p = 16 \) at 500 K. If initial pressures are \( P_{\ce{A}} = 2 \, \text{atm} \), \( P_{\ce{B}} = 1 \, \text{atm} \), what is \( P_{\ce{C}} \) at equilibrium?

Let \( P_{\ce{C}} = 2x \), \( P_{\ce{A}} = 2 - 2x \), \( P_{\ce{B}} = 1 - x \), total pressure = \( 3 - x \). \( K_p = \frac{(P_{\ce{C}})^2}{P_{\ce{A}}^2 P_{\ce{B}}} = \frac{(2x)^2}{(2 - 2x)^2 (1 - x)} = 16 \), \( \frac{4x^2}{4(1 - x)^2 (1 - x)} = 16 \), \( \frac{x^2}{(1 - x)^3} = 4 \), \( \frac{x}{1 - x} = 2 \), \( x = 2 - 2x \), \( 3x = 2 \), \( x = \frac{2}{3} \), \( P_{\ce{C}} = 2 \times \frac{2}{3} = 1.33 \, \text{atm} \).

2.0 atm
1.0 atm
1.33 atm
0.67 atm
3

The \( K_{sp} \) of \( \ce{Fe(OH)3} \) is \( 4.0 \times 10^{-38} \). What is the pH at which \( [\ce{Fe^{3+}}] = 1.0 \times 10^{-10} \, \text{M} \) in a saturated solution?

For \( \ce{Fe(OH)3 <=> Fe^{3+} + 3OH-} \), \( K_{sp} = [\ce{Fe^{3+}}][\ce{OH-}]^3 = 4.0 \times 10^{-38} \). Given \( [\ce{Fe^{3+}}] = 1.0 \times 10^{-10} \), \( (1.0 \times 10^{-10})[\ce{OH-}]^3 = 4.0 \times 10^{-38} \), \( [\ce{OH-}]^3 = 4.0 \times 10^{-28} \), \( [\ce{OH-}] = (4.0 \times 10^{-28})^{1/3} \approx 1.59 \times 10^{-9} \). \( \text{pOH} = -\log(1.59 \times 10^{-9}) \approx 8.8 \), \( \text{pH} = 14 - 8.8 = 5.2 \).

7.0
5.2
8.8
4.0
1

The \( K_{sp} \) of \( \ce{Mg(OH)2} \) is \( 1.8 \times 10^{-11} \). What is the solubility of \( \ce{Mg(OH)2} \) in pure water in mol/L?

For \( \ce{Mg(OH)2 <=> Mg^{2+} + 2OH-} \), \( K_{sp} = [\ce{Mg^{2+}}][\ce{OH-}]^2 = S (2S)^2 = 4S^3 = 1.8 \times 10^{-11} \). Solving, \( S^3 = \frac{1.8 \times 10^{-11}}{4} = 4.5 \times 10^{-12} \), \( S \approx 1.65 \times 10^{-4} \).

\( 1.65 \times 10^{-4} \)
\( 3.0 \times 10^{-4} \)
\( 4.5 \times 10^{-12} \)
\( 9.0 \times 10^{-6} \)
1

For \( \ce{2A(g) + B(g) <=> 3C(g)} \), \( K_p = 9 \) at 500 K. If \( P_{\ce{A}} = 0.5 \, \text{atm} \), \( P_{\ce{B}} = 0.5 \, \text{atm} \), and \( P_{\ce{C}} = 1.5 \, \text{atm} \) at equilibrium, what happens if pressure is doubled?

Initial: \( Q = \frac{(1.5)^3}{(0.5)^2 (0.5)} = 27 > K_p = 9 \), doubling pressure increases concentrations, \( Q \) increases further, shifts left.

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

For \( \ce{2NO(g) + Cl2(g) <=> 2NOCl(g)} \), \( K_p = 9 \) at 500 K. If initial pressures are \( P_{\ce{NO}} = 1 \, \text{atm} \), \( P_{\ce{Cl2}} = 0.5 \, \text{atm} \), what is \( P_{\ce{NOCl}} \) at equilibrium?

Let \( P_{\ce{NOCl}} = 2x \), \( P_{\ce{NO}} = 1 - 2x \), \( P_{\ce{Cl2}} = 0.5 - x \). \( K_p = \frac{(P_{\ce{NOCl}})^2}{(P_{\ce{NO}})^2 P_{\ce{Cl2}}} = \frac{(2x)^2}{(1 - 2x)^2 (0.5 - x)} = 9 \), \( \frac{4x^2}{(1 - 2x)^2 (0.5 - x)} = 9 \), \( x \approx 0.45 \), \( P_{\ce{NOCl}} = 0.9 \, \text{atm} \).

1.0 atm
0.9 atm
0.45 atm
1.8 atm
2

For the reaction \( \ce{CO(g) + Cl2(g) <=> COCl2(g)} \), if \( K_c = 9 \) and initial concentrations are \( [\ce{CO}] = 0.3 \, \text{M} \), \( [\ce{Cl2}] = 0.3 \, \text{M} \), what is \( [\ce{COCl2}] \) at equilibrium?

Let \( [\ce{COCl2}] = x \), \( [\ce{CO}] = 0.3 - x \), \( [\ce{Cl2}] = 0.3 - x \). \( K_c = \frac{[\ce{COCl2}]}{[\ce{CO}][\ce{Cl2}]} = \frac{x}{(0.3 - x)^2} = 9 \). Solving, \( \sqrt{\frac{x}{0.3 - x}} = 3 \), \( \frac{x}{0.3 - x} = 9 \), \( x = 2.7 - 9x \), \( 10x = 2.7 \), \( x = 0.27 \, \text{M} \).

0.3 M
0.15 M
0.27 M
0.03 M
3

In the equilibrium \( \ce{A(g) <=> 2B(g)} \), what happens if \( \ce{B} \) is removed at constant volume?

Removing \( \ce{B} \) decreases its concentration, shifting the equilibrium right to produce more \( \ce{B} \), per Le Chatelier’s principle.

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

The \( K_w \) of water at 373 K is \( 1.0 \times 10^{-12} \). What is the pH of pure water at this temperature?

\( K_w = [\ce{H+}][\ce{OH-}] = 1.0 \times 10^{-12} \), \( [\ce{H+}] = \sqrt{1.0 \times 10^{-12}} = 1.0 \times 10^{-6} \), \( \text{pH} = -\log(1.0 \times 10^{-6}) = 6 \).

7
12
5
6
4

A weak base \( \ce{BOH} \) (\( K_b = 4.0 \times 10^{-5} \)) has a 0.05 M solution with 2% ionization. What is the pH?

\( \alpha = 0.02 \), \( [\ce{OH-}] = 0.05 \times 0.02 = 1.0 \times 10^{-3} \), \( \text{pOH} = 3 \), \( \text{pH} = 14 - 3 = 11 \).

3.0
11.0
10.5
12.0
2

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