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  4. Chemical Equilibrium Chapter-Wise Test 2

Chemical Equilibrium Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

For \( \ce{HA <=> H+ + A-} \), if \( K_a = 1.0 \times 10^{-5} \) and \( [\ce{HA}] = 0.05 \, \text{M} \) at equilibrium, what is \( [\ce{H+}] \)?

\( K_a = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} = \frac{x^2}{0.05 - x} \approx \frac{x^2}{0.05} = 1.0 \times 10^{-5} \), \( x^2 = 5.0 \times 10^{-7} \), \( x = 7.07 \times 10^{-4} \, \text{M} \).

\( 1.0 \times 10^{-5} \)
\( 7.07 \times 10^{-4} \)
\( 5.0 \times 10^{-7} \)
\( 2.24 \times 10^{-3} \)
2

A weak acid \( \ce{HX} \) (\( K_a = 2.0 \times 10^{-5} \)) is mixed with 0.06 M \( \ce{NaX} \) in a 1:2 volume ratio (acid:salt). If \( [\ce{HX}] = 0.04 \, \text{M} \) after mixing, what is the pH?

Total volume = 3V, \( [\ce{X-}] = \frac{0.06 \times 2V}{3V} = 0.04 \, \text{M} \), \( \text{p}K_a = 4.7 \), \( \text{pH} = 4.7 + \log \frac{0.04}{0.04} = 4.7 \).

5.0
4.0
4.7
5.3
3

A weak acid \( \ce{HY} \) (\( K_a = 1.0 \times 10^{-4} \)) is mixed with its salt \( \ce{NaY} \) in a 2:1 molar ratio. If \( [\ce{HY}] = 0.2 \, \text{M} \), what is the pH?

\( [\ce{Y-}] = 0.1 \, \text{M} \), \( \text{p}K_a = 4 \), \( \text{pH} = \text{p}K_a + \log \frac{[\ce{Y-}]}{[\ce{HY}]} = 4 + \log \frac{0.1}{0.2} = 4 + \log 0.5 = 4 - 0.301 = 3.7 \).

4.0
4.3
3.0
3.7
4

Which of the following is a Bronsted-Lowry acid?

A Bronsted-Lowry acid donates a proton. \( \ce{HCO3-} \) can donate \( \ce{H+} \) to form \( \ce{CO3^{2-}} \).

\( \ce{OH-} \)
\( \ce{NH3} \)
\( \ce{HCO3-} \)
\( \ce{Cl-} \)
3

For the reaction \( \ce{H2(g) + I2(g) <=> 2HI(g)} \), the equilibrium concentrations are \( [\ce{H2}] = 0.02 \, \text{M} \), \( [\ce{I2}] = 0.02 \, \text{M} \), and \( [\ce{HI}] = 0.16 \, \text{M} \). What is the value of the equilibrium constant \( K_c \)?

\( K_c = \frac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]} = \frac{(0.16)^2}{(0.02)(0.02)} = \frac{0.0256}{0.0004} = 64 \).

64
32
16
8
1

For \( \ce{N2(g) + 3H2(g) <=> 2NH3(g)} \), \( K_p = 4.0 \times 10^{-3} \) at 600 K. If 1 mole of \( \ce{N2} \) and 3 moles of \( \ce{H2} \) are in a 2 L vessel, what is \( P_{\ce{NH3}} \) at equilibrium (\( R = 0.0831 \, \text{bar L/mol K} \))?

Initial: \( P_{\ce{N2}} = \frac{1 \times 0.0831 \times 600}{2} = 24.93 \, \text{bar} \), \( P_{\ce{H2}} = 74.79 \, \text{bar} \). Let \( 2x \) be \( P_{\ce{NH3}} \), \( P_{\ce{N2}} = 24.93 - x \), \( P_{\ce{H2}} = 74.79 - 3x \). \( K_p = \frac{(P_{\ce{NH3}})^2}{P_{\ce{N2}} (P_{\ce{H2}})^3} = \frac{(2x)^2}{(24.93 - x)(74.79 - 3x)^3} = 4.0 \times 10^{-3} \). Solving, \( x \approx 0.8 \), \( P_{\ce{NH3}} = 2 \times 0.8 = 1.6 \, \text{bar} \).

1.6 bar
2.4 bar
0.8 bar
3.2 bar
1

For \( \ce{2A(g) <=> B(g) + C(g)} \), \( K_p = 0.5 \) at 600 K. If the initial pressure of \( \ce{A} \) is 2 atm, what is \( P_{\ce{B}} \) at equilibrium?

Let \( P_{\ce{B}} = P_{\ce{C}} = x \), \( P_{\ce{A}} = 2 - 2x \), total pressure = \( 2 - 2x + 2x = 2 \). \( K_p = \frac{P_{\ce{B}} P_{\ce{C}}}{(P_{\ce{A}})^2} = \frac{x^2}{(2 - 2x)^2} = 0.5 \), \( \frac{x}{2 - 2x} = \sqrt{0.5} \approx 0.707 \), \( x \approx 0.828 \, \text{atm} \).

1.0 atm
0.828 atm
0.5 atm
2.0 atm
2

For \( \ce{2X(g) <=> Y(g)} \), if \( K_p = 0.25 \) at 300 K (\( R = 0.0831 \, \text{bar L/mol K} \)), what is \( K_c \)?

\( K_p = K_c (RT)^{\Delta n} \), \( \Delta n = 1 - 2 = -1 \), \( RT = 0.0831 \times 300 = 24.93 \). Thus, \( K_c = K_p \times RT = 0.25 \times 24.93 \approx 6.23 \).

6.23
0.25
0.01
24.93
1

For the reaction \( \ce{X(g) + Y(g) <=> 2Z(g)} \), if \( K_c = 25 \) and at equilibrium \( [\ce{Z}] = 0.5 \, \text{M} \), what is \( [\ce{X}] \) if \( [\ce{Y}] = 0.02 \, \text{M} \)?

\( K_c = \frac{[\ce{Z}]^2}{[\ce{X}][\ce{Y}]} = 25 \). Substituting, \( 25 = \frac{(0.5)^2}{[\ce{X}](0.02)} = \frac{0.25}{0.02 [\ce{X}]} \), \( 25 \times 0.02 [\ce{X}] = 0.25 \), \( 0.5 [\ce{X}] = 0.25 \), \( [\ce{X}] = 0.5 \, \text{M} \).

0.5 M
0.25 M
0.1 M
0.02 M
1

In a physical equilibrium \( \ce{H2O(l) <=> H2O(g)} \), what happens if the temperature is increased?

Increasing temperature favors the endothermic process (evaporation), shifting the equilibrium towards the vapor phase.

More vapor forms
More liquid forms
No change
Pressure decreases
1

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