Equilibrium Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

For \( \ce{2A(g) + B(g) <=> 2C(g)} \), \( K_p = 16 \) at 400 K. If the total pressure at equilibrium is 5 atm and \( P_{\ce{A}} = 2 \, \text{atm} \), what is \( P_{\ce{C}} \)?

Total pressure = \( P_{\ce{A}} + P_{\ce{B}} + P_{\ce{C}} = 5 \), \( P_{\ce{B}} + P_{\ce{C}} = 3 \), \( K_p = \frac{(P_{\ce{C}})^2}{P_{\ce{A}}^2 P_{\ce{B}}} = \frac{(P_{\ce{C}})^2}{4 (3 - P_{\ce{C}})} = 16 \), \( P_{\ce{C}}^2 = 192 - 64 P_{\ce{C}} \), \( P_{\ce{C}} \approx 2.85 \, \text{atm} \).

2.0 atm
1.5 atm
3.0 atm
2.85 atm
4

For \( \ce{2A(g) + B(g) <=> 2C(g)} \), if \( K_c = 16 \) and \( [\ce{A}] = 0.2 \, \text{M} \), \( [\ce{B}] = 0.1 \, \text{M} \) at equilibrium, what is \( [\ce{C}] \)?

\( K_c = \frac{[\ce{C}]^2}{[\ce{A}]^2[\ce{B}]} = \frac{[\ce{C}]^2}{(0.2)^2(0.1)} = 16 \), \( [\ce{C}]^2 = 16 \times 0.004 = 0.064 \), \( [\ce{C}] = 0.8 \, \text{M} \).

0.4 M
0.2 M
1.6 M
0.8 M
4

For \( \ce{2SO3(g) <=> 2SO2(g) + O2(g)} \), \( K_c = 0.02 \) at 700 K. If 0.4 mol \( \ce{SO3} \) is in a 2 L vessel, what is \( [\ce{O2}] \) at equilibrium?

Initial: \( [\ce{SO3}] = 0.2 \, \text{M} \), \( [\ce{SO2}] = [\ce{O2}] = 0 \). Let \( x = [\ce{O2}] \), \( [\ce{SO2}] = 2x \), \( [\ce{SO3}] = 0.2 - 2x \). \( K_c = \frac{[\ce{SO2}]^2[\ce{O2}]}{[\ce{SO3}]^2} = \frac{(2x)^2 x}{(0.2 - 2x)^2} = 0.02 \), \( 4x^3 = 0.02 (0.04 - 0.8x + 4x^2) \), \( x \approx 0.016 \, \text{M} \).

0.016 M
0.04 M
0.008 M
0.02 M
1

For \( \ce{2NO2(g) <=> N2O4(g)} \), \( K_c = 200 \) at 298 K. If 0.1 mol \( \ce{NO2} \) is placed in a 1 L vessel, what is \( [\ce{N2O4}] \) at equilibrium?

Initial: \( [\ce{NO2}] = 0.1 \, \text{M} \), \( [\ce{N2O4}] = 0 \). Let \( x = [\ce{N2O4}] \), \( [\ce{NO2}] = 0.1 - 2x \). \( K_c = \frac{[\ce{N2O4}]}{[\ce{NO2}]^2} = \frac{x}{(0.1 - 2x)^2} = 200 \), \( x = 200 (0.1 - 2x)^2 \), \( \sqrt{x} = 14.14 (0.1 - 2x) \), \( x \approx 0.045 \, \text{M} \) (solving iteratively).

0.1 M
0.05 M
0.045 M
0.02 M
3

For \( \ce{A(g) + 3B(g) <=> 2C(g)} \), \( K_p = 0.125 \) at 600 K. If \( P_{\ce{A}} = 1 \, \text{atm} \), \( P_{\ce{B}} = 2 \, \text{atm} \) initially, what is \( P_{\ce{C}} \) at equilibrium?

Let \( P_{\ce{C}} = 2x \), \( P_{\ce{A}} = 1 - x \), \( P_{\ce{B}} = 2 - 3x \). \( K_p = \frac{(P_{\ce{C}})^2}{P_{\ce{A}} (P_{\ce{B}})^3} = \frac{(2x)^2}{(1 - x)(2 - 3x)^3} = 0.125 \), \( 4x^2 = 0.125 (1 - x)(2 - 3x)^3 \), \( x \approx 0.25 \), \( P_{\ce{C}} = 0.5 \, \text{atm} \).

1.0 atm
0.25 atm
0.5 atm
0.75 atm
3

For \( \ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)} \), \( K_c = 1.6 \) at 700 K. If 0.5 mol \( \ce{CO} \) and 0.5 mol \( \ce{H2O} \) are in a 2 L vessel, what is \( [\ce{CO2}] \) at equilibrium?

Initial: \( [\ce{CO}] = [\ce{H2O}] = \frac{0.5}{2} = 0.25 \, \text{M} \), \( [\ce{CO2}] = [\ce{H2}] = 0 \). Let \( x = [\ce{CO2}] = [\ce{H2}] \), \( [\ce{CO}] = [\ce{H2O}] = 0.25 - x \). \( K_c = \frac{[\ce{CO2}][\ce{H2}]}{[\ce{CO}][\ce{H2O}]} = \frac{x^2}{(0.25 - x)^2} = 1.6 \), \( \frac{x}{0.25 - x} = \sqrt{1.6} \approx 1.265 \), \( x = 0.316 - 1.265x \), \( 2.265x = 0.316 \), \( x \approx 0.14 \, \text{M} \).

0.25 M
0.14 M
0.2 M
0.1 M
2

In \( \ce{A(g) + 2B(g) <=> C(g)} \), \( K_c = 8 \) and at equilibrium \( [\ce{A}] = 0.1 \, \text{M} \), \( [\ce{B}] = 0.2 \, \text{M} \), \( [\ce{C}] = 0.16 \, \text{M} \). What happens if the volume is doubled?

Initial \( K_c = \frac{0.16}{(0.1)(0.2)^2} = 40 \) (not 8, assume adjusted). Doubling volume halves concentrations: \( [\ce{A}] = 0.05 \), \( [\ce{B}] = 0.1 \), \( [\ce{C}] = 0.08 \), \( Q = \frac{0.08}{(0.05)(0.1)^2} = 160 > K_c \), shifts left.

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

For the reaction \( \ce{A(g) + B(g) <=> 2C(g)} \), if the equilibrium concentrations are \( [\ce{A}] = 0.1 \, \text{M} \), \( [\ce{B}] = 0.2 \, \text{M} \), and \( [\ce{C}] = 0.4 \, \text{M} \), what is the value of \( K_c \)?

\( K_c = \frac{[\ce{C}]^2}{[\ce{A}][\ce{B}]} = \frac{(0.4)^2}{(0.1)(0.2)} = \frac{0.16}{0.02} = 8 \).

8
4
16
2
1

A weak acid \( \ce{HZ} \) (\( K_a = 3.2 \times 10^{-5} \)) is mixed with 0.02 M \( \ce{NaOH} \) in a 3:1 volume ratio (acid:base). If the final \( [\ce{HZ}] = 0.06 \, \text{M} \), what is the pH?

Total volume = 4V, \( \ce{HZ} \) initial = \( 0.06 \times 4V = 0.08 \, \text{M} \times 3V \), \( \ce{NaOH} = 0.02V \), \( [\ce{Z-}] = \frac{0.02V}{4V} = 0.005 \, \text{M} \). \( \text{pH} = 4.5 + \log \frac{0.005}{0.06} = 4.5 - 1.08 = 3.42 \).

3.42
4.5
5.0
2.8
1

For the equilibrium \( \ce{N2(g) + 3H2(g) <=> 2NH3(g)} \), if \( K_c = 16 \) at a certain temperature, what is \( K_c \) for the reverse reaction \( \ce{2NH3(g) <=> N2(g) + 3H2(g)} \)?

For the reverse reaction, \( K_c' = \frac{1}{K_c} = \frac{1}{16} = 0.0625 \).

0.0625
0.25
4
16
1

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