Chemical Equilibrium Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

In \( \ce{A(g) + 2B(g) <=> 2C(g)} \), \( K_c = 4 \) and \( [\ce{A}] = 0.1 \, \text{M} \), \( [\ce{B}] = 0.2 \, \text{M} \), \( [\ce{C}] = 0.4 \, \text{M} \) at equilibrium. What happens if temperature increases (exothermic reaction)?

For an exothermic reaction, increasing temperature shifts equilibrium left, favoring reactants.

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

The \( K_a \) of \( \ce{HF} \) is \( 6.8 \times 10^{-4} \). What is the \( K_b \) of \( \ce{F-} \) at 298 K?

\( K_w = K_a \times K_b = 1.0 \times 10^{-14} \), \( K_b = \frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}} \approx 1.47 \times 10^{-11} \).

\( 6.8 \times 10^{-4} \)
\( 1.0 \times 10^{-14} \)
\( 3.4 \times 10^{-11} \)
\( 1.47 \times 10^{-11} \)
4

For the equilibrium \( \ce{2NO(g) <=> N2(g) + O2(g)} \), if \( K_c = 4 \) and initial concentration of \( \ce{NO} \) is 0.4 M with no products, what is \( [\ce{N2}] \) at equilibrium?

Let \( [\ce{N2}] = [\ce{O2}] = x \), \( [\ce{NO}] = 0.4 - 2x \). \( K_c = \frac{[\ce{N2}][\ce{O2}]}{[\ce{NO}]^2} = \frac{x^2}{(0.4 - 2x)^2} = 4 \). Taking square root, \( \frac{x}{0.4 - 2x} = 2 \), \( x = 0.8 - 4x \), \( 5x = 0.8 \), \( x = 0.16 \, \text{M} \).

0.4 M
0.16 M
0.08 M
0.2 M
2

For \( \ce{A2B(s) <=> 2A+(aq) + B^{2-}(aq)} \), if \( K_{sp} = 1.0 \times 10^{-9} \), what is the solubility of \( \ce{A2B} \) in mol/L?

\( K_{sp} = [\ce{A+}]^2[\ce{B^{2-}}] = (2S)^2 S = 4S^3 = 1.0 \times 10^{-9} \), \( S^3 = 2.5 \times 10^{-10} \), \( S \approx 6.3 \times 10^{-4} \).

\( 1.0 \times 10^{-3} \)
\( 6.3 \times 10^{-4} \)
\( 2.5 \times 10^{-10} \)
\( 5.0 \times 10^{-5} \)
2

The ionization constant of a weak acid \( \ce{HA} \) is \( 1.0 \times 10^{-5} \). What is the pH of a 0.1 M solution of this acid?

For \( \ce{HA <=> H+ + A-} \), \( K_a = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} = \frac{x^2}{0.1 - x} \approx \frac{x^2}{0.1} = 1.0 \times 10^{-5} \). Solving, \( x = \sqrt{1.0 \times 10^{-6}} = 1.0 \times 10^{-3} \), so \( \text{pH} = -\log(1.0 \times 10^{-3}) = 3 \).

3
2
4
5
1

Which of the following is a Lewis acid?

A Lewis acid accepts an electron pair. \( \ce{Fe^{3+}} \) has vacant orbitals and can accept an electron pair.

\( \ce{NH3} \)
\( \ce{OH-} \)
\( \ce{Cl-} \)
\( \ce{Fe^{3+}} \)
4

For the equilibrium \( \ce{A(g) <=> B(g) + C(g)} \), if the initial pressure of \( \ce{A} \) is 1 atm and at equilibrium the total pressure is 1.5 atm, what is \( K_p \)?

Let \( P_{\ce{B}} = P_{\ce{C}} = p \), \( P_{\ce{A}} = 1 - p \), total pressure = \( (1 - p) + p + p = 1 + p = 1.5 \), \( p = 0.5 \). \( P_{\ce{A}} = 0.5 \, \text{atm} \), \( P_{\ce{B}} = P_{\ce{C}} = 0.5 \, \text{atm} \). \( K_p = \frac{P_{\ce{B}} P_{\ce{C}}}{P_{\ce{A}}} = \frac{(0.5)(0.5)}{0.5} = 0.5 \).

1.0
0.5
2.0
0.25
2

For the reaction \( \ce{2HI(g) <=> H2(g) + I2(g)} \), if \( K_c = 0.04 \) at 700 K and \( [\ce{HI}] = 0.2 \, \text{M} \) at equilibrium, what is \( [\ce{H2}] \)?

\( K_c = \frac{[\ce{H2}][\ce{I2}]}{[\ce{HI}]^2} = 0.04 \). Since \( [\ce{H2}] = [\ce{I2}] = x \), \( 0.04 = \frac{x^2}{(0.2)^2} = \frac{x^2}{0.04} \), \( x^2 = 0.0016 \), \( x = 0.04 \, \text{M} \).

0.02 M
0.08 M
0.2 M
0.04 M
4

The \( K_{sp} \) of \( \ce{CuS} \) is \( 6.3 \times 10^{-36} \). What is the pH at which \( [\ce{Cu^{2+}}] = 1.0 \times 10^{-12} \, \text{M} \) in a saturated solution, given \( K_a \) of \( \ce{H2S} = 9.5 \times 10^{-8} \) and \( K_{a2} = 1.0 \times 10^{-19} \)?

For \( \ce{CuS <=> Cu^{2+} + S^{2-}} \), \( K_{sp} = [\ce{Cu^{2+}}][\ce{S^{2-}}] = 6.3 \times 10^{-36} \), \( [\ce{S^{2-}}] = 6.3 \times 10^{-24} \). For \( \ce{H2S <=> 2H+ + S^{2-}} \), \( K = K_{a1} \times K_{a2} = 9.5 \times 10^{-27} \), \( [\ce{S^{2-}}] = \frac{K [\ce{H2S}]}{[\ce{H+}]^2} \), assume \( [\ce{H2S}] = 0.1 \, \text{M} \), \( 6.3 \times 10^{-24} = \frac{9.5 \times 10^{-27} \times 0.1}{[\ce{H+}]^2} \), \( [\ce{H+}]^2 = 1.51 \times 10^{-4} \), \( [\ce{H+}] = 1.23 \times 10^{-2} \), \( \text{pH} \approx 1.91 \).

1.91
2.5
3.0
7.0
1

For \( \ce{Hg2Cl2(s) <=> Hg2^{2+}(aq) + 2Cl-(aq)} \), \( K_{sp} = 1.3 \times 10^{-18} \). What is \( [\ce{Hg2^{2+}}] \) in a 0.01 M \( \ce{NaCl} \) solution?

\( K_{sp} = [\ce{Hg2^{2+}}][\ce{Cl-}]^2 = 1.3 \times 10^{-18} \), \( [\ce{Cl-}] \approx 0.01 \), \( [\ce{Hg2^{2+}}] (0.01)^2 = 1.3 \times 10^{-18} \), \( [\ce{Hg2^{2+}}] = \frac{1.3 \times 10^{-18}}{0.0001} = 1.3 \times 10^{-14} \, \text{M} \).

\( 1.3 \times 10^{-18} \)
\( 1.3 \times 10^{-14} \)
\( 6.5 \times 10^{-15} \)
\( 2.6 \times 10^{-13} \)
2

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