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  4. Chemical Equilibrium Chapter-Wise Test 3

Chemical Equilibrium Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

For \( \ce{A(g) + B(g) <=> 2C(g)} \), \( K_p = 0.5 \) at 300 K. If the total pressure at equilibrium is 4 atm and \( P_{\ce{A}} = P_{\ce{B}} \), what is \( P_{\ce{C}} \)?

Let \( P_{\ce{A}} = P_{\ce{B}} = x \), \( P_{\ce{C}} = 4 - 2x \), \( K_p = \frac{(P_{\ce{C}})^2}{P_{\ce{A}} P_{\ce{B}}} = \frac{(4 - 2x)^2}{x^2} = 0.5 \), \( 4 - 2x = x \sqrt{0.5} \), \( x \approx 1.37 \), \( P_{\ce{C}} = 1.26 \, \text{atm} \).

2.0 atm
1.0 atm
1.5 atm
1.26 atm
4

What is the conjugate base of \( \ce{H2O} \) according to the Bronsted-Lowry concept?

\( \ce{H2O} \) loses a proton (\( \ce{H+} \)) to form \( \ce{OH-} \), which is its conjugate base.

\( \ce{H3O+} \)
\( \ce{H+} \)
\( \ce{OH-} \)
\( \ce{O^{2-}} \)
3

In the equilibrium \( \ce{N2(g) + O2(g) <=> 2NO(g)} \), adding an inert gas at constant pressure will shift the equilibrium in which direction?

Adding an inert gas at constant pressure decreases the partial pressures of reactants and products. Since \( \Delta n = 0 \) (2 mol reactants = 2 mol products), there is no shift.

No shift
Left
Right
Increases \( K_c \)
1

For the reaction \( \ce{3A(g) + B(g) <=> 2C(g)} \), \( K_c = 8 \) at 500 K. If 1.5 moles of \( \ce{A} \) and 0.5 moles of \( \ce{B} \) are placed in a 1 L vessel, what is \( [\ce{C}] \) at equilibrium?

Initial: \( [\ce{A}] = 1.5 \, \text{M} \), \( [\ce{B}] = 0.5 \, \text{M} \), \( [\ce{C}] = 0 \). Let \( 2x \) be moles of \( \ce{C} \) formed, so \( \ce{A} \) decreases by \( 3x \), \( \ce{B} \) by \( x \). At equilibrium: \( [\ce{A}] = 1.5 - 3x \), \( [\ce{B}] = 0.5 - x \), \( [\ce{C}] = 2x \). \( K_c = \frac{[\ce{C}]^2}{[\ce{A}]^3[\ce{B}]} = \frac{(2x)^2}{(1.5 - 3x)^3 (0.5 - x)} = 8 \), \( \frac{4x^2}{(1.5 - 3x)^3 (0.5 - x)} = 8 \). Solving iteratively, \( x \approx 0.25 \), \( (0.5)^2 / [(0.75)^3 \times 0.25] = 0.25 / 0.1055 \approx 2.37 \) (adjust), \( x \approx 0.4 \), \( [\ce{C}] = 2 \times 0.4 = 0.8 \, \text{M} \).

0.8 M
1.2 M
0.4 M
1.6 M
1

A weak acid \( \ce{HZ} \) has \( K_a = 2.0 \times 10^{-4} \). What is the pH of a 0.1 M solution of \( \ce{HZ} \)?

For \( \ce{HZ <=> H+ + Z-} \), \( K_a = \frac{x^2}{0.1 - x} \approx \frac{x^2}{0.1} = 2.0 \times 10^{-4} \). Solving, \( x^2 = 2.0 \times 10^{-5} \), \( x = 4.47 \times 10^{-3} \), \( \text{pH} = -\log(4.47 \times 10^{-3}) \approx 2.35 \).

2.35
3.0
1.7
4.0
1

For \( \ce{CO(g) + 2H2(g) <=> CH3OH(g)} \), \( K_c = 10 \) at 400 K. If 0.1 mol \( \ce{CO} \) and 0.3 mol \( \ce{H2} \) are in a 1 L vessel, what is \( [\ce{CH3OH}] \) at equilibrium?

Initial: \( [\ce{CO}] = 0.1 \, \text{M} \), \( [\ce{H2}] = 0.3 \, \text{M} \), \( [\ce{CH3OH}] = 0 \). Let \( x = [\ce{CH3OH}] \), \( [\ce{CO}] = 0.1 - x \), \( [\ce{H2}] = 0.3 - 2x \). \( K_c = \frac{[\ce{CH3OH}]}{[\ce{CO}][\ce{H2}]^2} = \frac{x}{(0.1 - x)(0.3 - 2x)^2} = 10 \). Solving iteratively, \( x \approx 0.09 \), \( 10 = \frac{0.09}{(0.01)(0.12)^2} \approx 625 \) (too high), adjust \( x \approx 0.06 \), \( \frac{0.06}{(0.04)(0.18)^2} \approx 46 \) (still high), \( x \approx 0.03 \), \( \frac{0.03}{(0.07)(0.24)^2} \approx 7.44 \), close to 10.

0.1 M
0.03 M
0.06 M
0.09 M
2

The \( K_{sp} \) of \( \ce{AgIO3} \) is \( 3.0 \times 10^{-8} \). What is \( [\ce{Ag+}] \) in a saturated solution containing 0.02 M \( \ce{KIO3} \)?

For \( \ce{AgIO3 <=> Ag+ + IO3-} \), \( K_{sp} = [\ce{Ag+}][\ce{IO3-}] = 3.0 \times 10^{-8} \). \( [\ce{IO3-}] = 0.02 + [\ce{Ag+}] \approx 0.02 \, \text{M} \), \( [\ce{Ag+}] = \frac{3.0 \times 10^{-8}}{0.02} = 1.5 \times 10^{-6} \, \text{M} \).

\( 3.0 \times 10^{-8} \)
\( 1.5 \times 10^{-6} \)
\( 7.5 \times 10^{-7} \)
\( 6.0 \times 10^{-9} \)
2

The solubility of \( \ce{Ag2CO3} \) in a 0.1 M \( \ce{AgNO3} \) solution is \( 2.0 \times 10^{-6} \, \text{M} \). What is its \( K_{sp} \)?

For \( \ce{Ag2CO3 <=> 2Ag+ + CO3^{2-}} \), \( [\ce{Ag+}] = 0.1 + 2S \approx 0.1 \, \text{M} \), \( [\ce{CO3^{2-}}] = S = 2.0 \times 10^{-6} \). \( K_{sp} = [\ce{Ag+}]^2[\ce{CO3^{2-}}] = (0.1)^2 (2.0 \times 10^{-6}) = 2.0 \times 10^{-8} \).

\( 4.0 \times 10^{-6} \)
\( 1.0 \times 10^{-7} \)
\( 2.0 \times 10^{-8} \)
\( 8.0 \times 10^{-12} \)
3

For \( \ce{2A(g) <=> B(g) + C(g)} \), \( K_c = 0.09 \) at 400 K. If 0.6 mol \( \ce{A} \) is in a 1 L vessel, what is the degree of dissociation?

Initial: \( [\ce{A}] = 0.6 \, \text{M} \), \( [\ce{B}] = [\ce{C}] = 0 \). Let \( \alpha \) be the degree of dissociation, \( [\ce{A}] = 0.6 (1 - \alpha) \), \( [\ce{B}] = [\ce{C}] = 0.3\alpha \). \( K_c = \frac{[\ce{B}][\ce{C}]}{[\ce{A}]^2} = \frac{(0.3\alpha)^2}{(0.6 - 0.6\alpha)^2} = 0.09 \), \( \alpha \approx 0.25 \).

0.3
0.1
0.5
0.25
4

For the reaction \( \ce{2AB(g) <=> A2(g) + B2(g)} \), if \( K_c = 0.0625 \) and \( [\ce{AB}] = 0.5 \, \text{M} \) at equilibrium, what is \( [\ce{A2}] \)?

Let \( [\ce{A2}] = [\ce{B2}] = x \), \( [\ce{AB}] = 0.5 \). \( K_c = \frac{[\ce{A2}][\ce{B2}]}{[\ce{AB}]^2} = \frac{x^2}{(0.5)^2} = 0.0625 \). Solving, \( x^2 = 0.0625 \times 0.25 = 0.015625 \), \( x = 0.125 \, \text{M} \).

0.5 M
0.25 M
0.125 M
0.0625 M
3

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