Equilibrium Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

For \( \ce{N2O4(g) <=> 2NO2(g)} \), \( K_p = 0.16 \) at 400 K. If 0.5 moles of \( \ce{N2O4} \) are placed in a 1 L vessel, what is \( P_{\ce{NO2}} \) at equilibrium (\( R = 0.0831 \, \text{bar L/mol K} \))?

Initial: \( P_{\ce{N2O4}} = \frac{0.5 \times 0.0831 \times 400}{1} = 16.62 \, \text{bar} \). Let \( 2x = P_{\ce{NO2}} \), \( P_{\ce{N2O4}} = 16.62 - x \), total pressure = \( 16.62 + x \). \( K_p = \frac{(P_{\ce{NO2}})^2}{P_{\ce{N2O4}}} = \frac{(2x)^2}{16.62 - x} = 0.16 \), \( 4x^2 = 0.16 (16.62 - x) \), \( x \approx 0.8 \), \( P_{\ce{NO2}} = 1.6 \, \text{bar} \).

1.6 bar
2.0 bar
0.8 bar
3.2 bar
1

Which species can act as both a Lewis acid and base?

\( \ce{H2O} \) can accept an electron pair (Lewis acid) or donate a lone pair (Lewis base), making it amphoteric.

\( \ce{BF3} \)
\( \ce{NH3} \)
\( \ce{H2O} \)
\( \ce{H+} \)
3

For \( \ce{2A(g) + B(g) <=> 2C(g)} \), \( K_p = 4 \) at 500 K. If initial pressures are \( P_{\ce{A}} = 2 \, \text{atm} \), \( P_{\ce{B}} = 1 \, \text{atm} \), what is \( P_{\ce{C}} \) at equilibrium?

Let \( P_{\ce{C}} = 2x \), \( P_{\ce{A}} = 2 - 2x \), \( P_{\ce{B}} = 1 - x \). \( K_p = \frac{(P_{\ce{C}})^2}{P_{\ce{A}}^2 P_{\ce{B}}} = \frac{(2x)^2}{(2 - 2x)^2 (1 - x)} = 4 \), \( \frac{4x^2}{(4 - 4x + 4x^2)(1 - x)} = 4 \), \( x \approx 0.8 \), \( P_{\ce{C}} = 1.6 \, \text{atm} \).

2.0 atm
1.6 atm
0.8 atm
1.2 atm
2

What is the \( \text{pOH} \) of a 0.001 M solution of \( \ce{NaOH} \) assuming complete dissociation?

For \( \ce{NaOH} \), \( [\ce{OH-}] = 0.001 \, \text{M} = 10^{-3} \, \text{M} \), so \( \text{pOH} = -\log(10^{-3}) = 3 \).

2
3
11
14
2

The \( K_a \) of \( \ce{HCN} \) is \( 4.9 \times 10^{-10} \). What is \( K_b \) of \( \ce{CN-} \) at 298 K?

\( K_w = K_a \times K_b = 1.0 \times 10^{-14} \), so \( K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{4.9 \times 10^{-10}} \approx 2.04 \times 10^{-5} \).

\( 4.9 \times 10^{-10} \)
\( 1.0 \times 10^{-14} \)
\( 5.0 \times 10^{-5} \)
\( 2.04 \times 10^{-5} \)
4

The solubility of \( \ce{Ag2CrO4} \) in 0.02 M \( \ce{K2CrO4} \) is \( 1.0 \times 10^{-5} \, \text{M} \). What is its \( K_{sp} \)?

For \( \ce{Ag2CrO4 <=> 2Ag+ + CrO4^{2-}} \), \( [\ce{Ag+}] = 2.0 \times 10^{-5} \), \( [\ce{CrO4^{2-}}] \approx 0.02 \), \( K_{sp} = (2.0 \times 10^{-5})^2 \times 0.02 = 8.0 \times 10^{-12} \).

\( 1.0 \times 10^{-5} \)
\( 4.0 \times 10^{-10} \)
\( 8.0 \times 10^{-12} \)
\( 2.0 \times 10^{-11} \)
3

Which species can act as both a Bronsted-Lowry acid and base?

\( \ce{HS-} \) can donate \( \ce{H+} \) (acid) to form \( \ce{S^{2-}} \) or accept \( \ce{H+} \) (base) to form \( \ce{H2S} \), making it amphoteric.

\( \ce{HCl} \)
\( \ce{OH-} \)
\( \ce{HS-} \)
\( \ce{Na+} \)
3

A weak acid \( \ce{HA} \) (\( K_a = 5.0 \times 10^{-6} \)) is mixed with 0.04 M \( \ce{NaOH} \) in a 2:1 volume ratio (acid:base). If the final \( [\ce{HA}] = 0.06 \, \text{M} \), what is the pH?

Total volume = 3V, initial \( [\ce{HA}] = 0.09 \, \text{M} \), \( [\ce{A-}] = \frac{0.04V}{3V} = 0.0133 \, \text{M} \), \( \text{p}K_a = 5.3 \), \( \text{pH} = 5.3 + \log \frac{0.0133}{0.06} = 5.3 - 0.655 = 4.65 \).

4.65
5.3
6.0
4.0
1

For \( \ce{XY2(s) <=> X^{2+}(aq) + 2Y-(aq)} \), if \( K_{sp} = 2.7 \times 10^{-7} \), what is the solubility of \( \ce{XY2} \) in mol/L?

\( K_{sp} = [\ce{X^{2+}}][\ce{Y-}]^2 = S (2S)^2 = 4S^3 = 2.7 \times 10^{-7} \), \( S^3 = \frac{2.7 \times 10^{-7}}{4} = 6.75 \times 10^{-8} \), \( S \approx 4.07 \times 10^{-3} \).

\( 2.7 \times 10^{-7} \)
\( 4.07 \times 10^{-3} \)
\( 1.5 \times 10^{-3} \)
\( 6.75 \times 10^{-8} \)
2

Which species acts as a Bronsted-Lowry base?

\( \ce{HCO3-} \) can accept \( \ce{H+} \) to form \( \ce{H2CO3} \), making it a base.

\( \ce{H3O+} \)
\( \ce{HCl} \)
\( \ce{HCO3-} \)
\( \ce{H2SO4} \)
3

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