Equilibrium Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

For \( \ce{2SO3(g) <=> 2SO2(g) + O2(g)} \), \( K_p = 0.01 \) at 900 K. If 1 mole of \( \ce{SO3} \) is placed in a 1 L vessel, what is the partial pressure of \( \ce{O2} \) at equilibrium (\( R = 0.0831 \, \text{bar L/mol K} \))?

Initial: \( P_{\ce{SO3}} = \frac{1 \times 0.0831 \times 900}{1} = 74.79 \, \text{bar} \). Let \( 2x \) mol of \( \ce{SO3} \) dissociate, so \( P_{\ce{SO3}} = 74.79(1 - 2x) \), \( P_{\ce{SO2}} = 74.79 \times 2x \), \( P_{\ce{O2}} = 74.79x \), total pressure = 74.79. \( K_p = \frac{(P_{\ce{SO2}})^2 P_{\ce{O2}}}{(P_{\ce{SO3}})^2} = \frac{(74.79 \times 2x)^2 (74.79x)}{[74.79(1 - 2x)]^2} = 0.01 \). Simplifying, \( \frac{4x^2 \cdot 74.79x}{(1 - 2x)^2} = 0.01 \), \( 299.16x^3 = 0.01 (1 - 2x)^2 \). Solving, \( x \approx 0.013 \), \( P_{\ce{O2}} = 74.79 \times 0.013 \approx 0.972 \, \text{bar} \).

0.972 bar
1.5 bar
0.5 bar
2.0 bar
1

Which factor does not affect the position of equilibrium in \( \ce{A(g) + B(g) <=> C(g)} \)?

Adding a catalyst increases the rate of both forward and reverse reactions equally, thus not affecting the equilibrium position.

Pressure
Temperature
Concentration
Catalyst
4

In the equilibrium \( \ce{A2(g) <=> 2A(g)} \), if \( K_p = 0.5 \) at 400 K, what happens when the volume is doubled at constant temperature?

Decreasing pressure (doubling volume) favors the side with more moles of gas. Here, \( \Delta n = 2 - 1 = 1 \), so the equilibrium shifts to the right (towards \( \ce{A} \)).

Shifts left
Shifts right
No shift
\( K_p \) increases
2

For the reaction \( \ce{PCl5(g) <=> PCl3(g) + Cl2(g)} \), if \( K_p = 1.8 \) at 500 K and \( R = 0.0831 \, \text{bar L/mol K} \), what is \( K_c \)?

\( K_p = K_c (RT)^{\Delta n} \), where \( \Delta n = 2 - 1 = 1 \). Thus, \( K_c = \frac{K_p}{RT} = \frac{1.8}{0.0831 \times 500} = \frac{1.8}{41.55} \approx 0.043 \).

1.8
0.043
0.086
41.55
2

The \( K_a \) of \( \ce{HNO2} \) is \( 4.5 \times 10^{-4} \). What is the \( K_b \) of \( \ce{NO2-} \) at 298 K?

\( K_w = K_a \times K_b = 1.0 \times 10^{-14} \), \( K_b = \frac{1.0 \times 10^{-14}}{4.5 \times 10^{-4}} \approx 2.22 \times 10^{-11} \).

\( 4.5 \times 10^{-4} \)
\( 1.0 \times 10^{-14} \)
\( 5.0 \times 10^{-10} \)
\( 2.22 \times 10^{-11} \)
4

A solution of \( \ce{HA} \) (\( K_a = 1.0 \times 10^{-6} \)) has a pH of 4. What is the concentration of \( \ce{HA} \)?

\( \text{pH} = 4 \), \( [\ce{H+}] = 10^{-4} \). For \( \ce{HA <=> H+ + A-} \), \( K_a = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} = \frac{(10^{-4})^2}{[\ce{HA}]} = 1.0 \times 10^{-6} \). Solving, \( [\ce{HA}] = \frac{10^{-8}}{10^{-6}} = 0.01 \, \text{M} \).

0.1 M
1.0 M
0.01 M
0.001 M
3

For \( \ce{2P(g) <=> Q(g)} \), if \( K_p = 0.1 \) at 500 K (\( R = 0.0831 \, \text{bar L/mol K} \)), what is \( K_c \)?

\( K_p = K_c (RT)^{\Delta n} \), \( \Delta n = 1 - 2 = -1 \), \( RT = 0.0831 \times 500 = 41.55 \). Thus, \( K_c = K_p \times RT = 0.1 \times 41.55 = 4.155 \).

4.155
0.1
0.0024
41.55
1

For \( \ce{A(g) <=> 2B(g)} \), \( K_c = 0.16 \) at 400 K. If 0.5 mol \( \ce{A} \) is in a 1 L vessel, what is \( [\ce{B}] \) at equilibrium?

Initial: \( [\ce{A}] = 0.5 \, \text{M} \), \( [\ce{B}] = 0 \). Let \( 2x = [\ce{B}] \), \( [\ce{A}] = 0.5 - x \). \( K_c = \frac{[\ce{B}]^2}{[\ce{A}]} = \frac{(2x)^2}{0.5 - x} = 0.16 \), \( 4x^2 = 0.08 - 0.16x \), \( x \approx 0.1 \), \( [\ce{B}] = 0.2 \, \text{M} \).

0.5 M
0.1 M
0.2 M
0.4 M
3

For \( \ce{2A(g) <=> B(g)} \), \( K_c = 0.25 \) at 500 K. If 0.4 mol \( \ce{A} \) is in a 2 L vessel, what is the degree of dissociation?

Initial: \( [\ce{A}] = 0.2 \, \text{M} \), \( [\ce{B}] = 0 \). Let \( \alpha \) be the degree of dissociation, \( [\ce{A}] = 0.2 (1 - \alpha) \), \( [\ce{B}] = 0.1\alpha \). \( K_c = \frac{[\ce{B}]}{[\ce{A}]^2} = \frac{0.1\alpha}{(0.2 - 0.2\alpha)^2} = 0.25 \), \( \alpha \approx 0.36 \).

0.2
0.5
0.1
0.36
4

The \( K_{sp} \) of \( \ce{CaCO3} \) is \( 3.8 \times 10^{-9} \). What is its solubility in pure water in mol/L?

For \( \ce{CaCO3 <=> Ca^{2+} + CO3^{2-}} \), \( K_{sp} = [\ce{Ca^{2+}}][\ce{CO3^{2-}}] = S^2 = 3.8 \times 10^{-9} \), \( S = \sqrt{3.8 \times 10^{-9}} \approx 6.16 \times 10^{-5} \).

\( 6.16 \times 10^{-5} \)
\( 1.9 \times 10^{-4} \)
\( 3.8 \times 10^{-9} \)
\( 9.0 \times 10^{-5} \)
1

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0