Equilibrium Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

For \( \ce{N2O4(g) <=> 2NO2(g)} \), \( K_p = 0.16 \) at 298 K. If the initial pressure of \( \ce{N2O4} \) is 2 atm in a closed vessel, what is the total pressure at equilibrium (\( R = 0.0831 \, \text{bar L/mol K} \))?

Let \( P_{\ce{NO2}} = 2x \), \( P_{\ce{N2O4}} = 2 - x \), total pressure = \( 2 - x + 2x = 2 + x \). \( K_p = \frac{(P_{\ce{NO2}})^2}{P_{\ce{N2O4}}} = \frac{(2x)^2}{2 - x} = 0.16 \), \( \frac{4x^2}{2 - x} = 0.16 \), \( 4x^2 = 0.32 - 0.16x \), \( 4x^2 + 0.16x - 0.32 = 0 \), \( x^2 + 0.04x - 0.08 = 0 \), \( x = \frac{-0.04 \pm \sqrt{0.0016 + 0.32}}{2} \), \( x \approx 0.27 \). Total pressure = \( 2 + 0.27 = 2.27 \, \text{atm} \).

2.0 atm
2.27 atm
2.54 atm
3.0 atm
2

For \( \ce{H2(g) + I2(g) <=> 2HI(g)} \), \( K_c = 50 \) at 700 K. If 0.2 mol \( \ce{H2} \) and 0.3 mol \( \ce{I2} \) are in a 1 L vessel, what is \( [\ce{HI}] \) at equilibrium?

Initial: \( [\ce{H2}] = 0.2 \, \text{M} \), \( [\ce{I2}] = 0.3 \, \text{M} \), \( [\ce{HI}] = 0 \). Let \( 2x = [\ce{HI}] \), \( [\ce{H2}] = 0.2 - x \), \( [\ce{I2}] = 0.3 - x \). \( K_c = \frac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]} = \frac{(2x)^2}{(0.2 - x)(0.3 - x)} = 50 \), \( 4x^2 = 50 (0.06 - 0.5x + x^2) \), \( 46x^2 - 25x + 3 = 0 \), \( x \approx 0.15 \), \( [\ce{HI}] = 0.3 \, \text{M} \).

0.2 M
0.3 M
0.4 M
0.5 M
2

In \( \ce{2A(g) <=> B(g) + C(g)} \), \( K_c = 0.01 \) and \( [\ce{A}] = 0.2 \, \text{M} \), \( [\ce{B}] = 0.02 \, \text{M} \), \( [\ce{C}] = 0.02 \, \text{M} \) at equilibrium. What happens if \( \ce{B} \) is removed?

Removing \( \ce{B} \) decreases \( Q \), shifts right to produce more \( \ce{B} \) and \( \ce{C} \).

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

For \( \ce{A(g) + 2B(g) <=> 2C(g)} \), \( K_p = 1 \) at 400 K. If \( P_{\ce{A}} = 0.5 \, \text{atm} \), \( P_{\ce{B}} = 1 \, \text{atm} \) initially, what is \( P_{\ce{C}} \) at equilibrium?

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

1.0 atm
0.25 atm
0.5 atm
0.75 atm
3

The \( K_b \) of \( \ce{NH3} \) is \( 1.8 \times 10^{-5} \). What is the \( K_a \) of \( \ce{NH4+} \)?

\( K_w = K_a \times K_b = 1.0 \times 10^{-14} \), \( K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} \).

\( 1.8 \times 10^{-5} \)
\( 1.0 \times 10^{-14} \)
\( 2.0 \times 10^{-9} \)
\( 5.56 \times 10^{-10} \)
4

For the reaction \( \ce{2SO2(g) + O2(g) <=> 2SO3(g)} \), if the initial moles of \( \ce{SO2} \) and \( \ce{O2} \) are 2 and 1 respectively in a 1 L vessel, and at equilibrium 1 mole of \( \ce{SO3} \) is formed, what is \( K_c \)?

At equilibrium: \( \ce{SO2} = 2 - 1 = 1 \, \text{M} \), \( \ce{O2} = 1 - 0.5 = 0.5 \, \text{M} \), \( \ce{SO3} = 1 \, \text{M} \). \( K_c = \frac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]} = \frac{(1)^2}{(1)^2(0.5)} = \frac{1}{0.5} = 2 \).

1
4
2
0.5
3

For \( \ce{2SO2(g) + O2(g) <=> 2SO3(g)} \), \( K_p = 100 \) at 800 K. If initial pressures are \( P_{\ce{SO2}} = 1 \, \text{atm} \), \( P_{\ce{O2}} = 0.5 \, \text{atm} \), what is \( P_{\ce{SO3}} \) at equilibrium?

Let \( P_{\ce{SO3}} = 2x \), \( P_{\ce{SO2}} = 1 - 2x \), \( P_{\ce{O2}} = 0.5 - x \). \( K_p = \frac{(P_{\ce{SO3}})^2}{P_{\ce{SO2}}^2 P_{\ce{O2}}} = \frac{(2x)^2}{(1 - 2x)^2 (0.5 - x)} = 100 \), \( \frac{4x^2}{(1 - 2x)^2 (0.5 - x)} = 100 \), \( \frac{2x}{1 - 2x} \cdot \frac{1}{\sqrt{0.5 - x}} = 10 \), \( x \approx 0.45 \), \( P_{\ce{SO3}} = 0.9 \, \text{atm} \).

1.0 atm
0.9 atm
0.45 atm
1.8 atm
2

For \( \ce{BaF2(s) <=> Ba^{2+}(aq) + 2F-(aq)} \), \( K_{sp} = 1.7 \times 10^{-6} \). In a 0.01 M \( \ce{NaF} \) solution, what is \( [\ce{Ba^{2+}}] \) in a saturated solution?

\( K_{sp} = [\ce{Ba^{2+}}][\ce{F-}]^2 = 1.7 \times 10^{-6} \). \( [\ce{F-}] = 0.01 + 2[\ce{Ba^{2+}}] \approx 0.01 \, \text{M} \) (since \( [\ce{Ba^{2+}} \) is small). \( [\ce{Ba^{2+}}] (0.01)^2 = 1.7 \times 10^{-6} \), \( [\ce{Ba^{2+}}] = \frac{1.7 \times 10^{-6}}{0.0001} = 1.7 \times 10^{-2} \, \text{M} \).

\( 1.7 \times 10^{-6} \)
\( 1.7 \times 10^{-2} \)
\( 4.25 \times 10^{-3} \)
\( 8.5 \times 10^{-4} \)
2

The equilibrium constant \( K_p \) for \( \ce{N2O4(g) <=> 2NO2(g)} \) is 0.98 at 298 K. What is \( K_c \) at this temperature (\( R = 0.0831 \, \text{bar L/mol K} \))?

\( K_p = K_c (RT)^{\Delta n} \), \( \Delta n = 2 - 1 = 1 \), \( RT = 0.0831 \times 298 \approx 24.76 \). Thus, \( K_c = \frac{K_p}{RT} = \frac{0.98}{24.76} \approx 0.04 \).

0.98
24.76
0.04
0.02
3

In the equilibrium \( \ce{2XY(g) <=> X2(g) + Y2(g)} \), decreasing the volume shifts the equilibrium in which direction?

Increasing pressure (decreasing volume) favors the side with fewer moles of gas. Here, \( 2 \, \text{mol} \rightarrow 2 \, \text{mol} \), so \( \Delta n = 0 \), and there is no shift.

Left
Right
No shift
Depends on \( K_c \)
3

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