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AP Chemistry : Reactions and Equilibrium

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Example Questions

Example Question #53 : Chemical Equilibrium

Calculate the equilibrium constant for $N_2 (g) + O_2 (g) + Cl_2 (g) \Leftrightarrow 2 \: NOCl$ given the following information:

$\frac{1}{2}N_2 + O_2(g) \Leftrightarrow NO_2 (g)$ $K_1 = 1.0 \, x \, 10^{-9}$

$NOCl(g) + \frac{1}{2} O_2 (g) \Leftrightarrow NO_2Cl$ $K_2 = 1.1 \, x\, 10^2$

$NO_2 (g) + \frac{1}{2} Cl_2 \Leftrightarrow NO_2Cl$ K_3 = 0.3

Possible Answers:

$7\, x\, 10^{-24}$

$7\, x\, 10^{-8}$

$3\, x\, 10^{-3}$

$5 \, x \, 10^{-10}$

Correct answer:

$7\, x\, 10^{-24}$

Explanation:

$2 (1/2 N_2 + O_2 \Leftrightarrow 2 NO_2 )$ $K = K_1^2 = 1 x 10^{-18}$

$2 (NO_2Cl \Leftrightarrow NOCl + 1/2 O_2)$ $K = (1/K_2)^2 = 8.3 x 10^{-5}$

$2 (NO_2 + 1/2 Cl_2 \Leftrightarrow NO_2Cl)$ K = K_3^2 = 0.09

$K = (1 x 10^{-18}) (8.3 x 10^{-5}) (0.09) = 7 x 10^{-24}$

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Example Question #54 : Chemical Equilibrium

Calculate the equilibrium constant for $2 NH_3 (g) + 3 I_2 (g) \Leftrightarrow N_2 (g) + 6 HI (g)$ given the following information:

$N_2 (g) + 3 H_2 (g) \Leftrightarrow 2 NH_3(g)$ K_1 = 0.50

$H_2 (g) + I_2 (g) \Leftrightarrow 2 HI (g)$ K_2 = 50

Possible Answers:

5.6 x 10^4

$3.2 x 10^{-3}$

1.5 x 10^3

2.8 x 10^6

2.5 x 10^5

Correct answer:

2.5 x 10^5

Explanation:

$2 NH_3 \Leftrightarrow N_2 + 3 H_2$ K = 1/K_1 = 2

$3 ( H_2 + I_2 \Leftrightarrow 2 HI)$ K = K_2^3 = 1.25 x 10^5

K = (2) (1.25 x 10^5) = 2.5 x 10^5

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Example Question #55 : Chemical Equilibrium

Calculate the equilibrium constant for $2 HI (g) \Leftrightarrow H_2 (g) + I_2 (g)$ given the following information:

$H_2(g) + I_2 (g) \Leftrightarrow 2 HI (g)$ K_1 = 50

Possible Answers:

1.21

0.25

0.05

0.01

0.02

Correct answer:

0.02

Explanation:

$2 HI \Leftrightarrow H_2 + I_2$ K = 1/K_1 = 0.02

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Example Question #1 : Calculating The Equilibrium Concentrations

Consider formation of Nitrogen monoxide: $N_2 (g) + O_2 (g) \Leftrightarrow 2 NO (g)$ with $K = 4.2 x 10^{-8}$ . If the initial concentration of N2 was 0.085M and O2 was 0.038 M, what is the concentration of nitrogen monoxide at equilibrium?

Possible Answers:

1.1 x 10^-3 M

3.2 x 10^-8 M

1.1 x 10^-4M

1.2 x 10^-6 M

1.2 x 10^-5 M

Correct answer:

1.2 x 10^-5 M

Explanation:

$K = \frac{[NO]^2}{[N_2 ][O_2 ]} =\frac{(2x)^2} {(0.085-x)(0.038-x) }$

$\sim \frac{4x^2}{(0.085)(0.038)} = 4.2 x 10^{-8}$

$x = 5.82 x 10^{-6}$

$[NO] = 2x = 2(6.82 x 10^{-6}) = 1.2 x 10^{-5}$

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Example Question #56 : Chemical Equilibrium

Given the ICE table below, what are the signs on the terms that would appear in the “Change” row?

$M + 2L \Leftrightarrow ML2$

Possible Answers:

-x, -2x, +x

+x, +4x, -x

-x, -x, +x

+x, +x, +x

+x, +2x, +x

Correct answer:

-x, -2x, +x

Explanation:

Based on the balanced equation, one would use -x, -2x, and +x.

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Example Question #141 : Reactions And Equilibrium

Find the equilibrium concentrations for C in the following chemical reaction: A + B -> 2C

K = 9.0 x 10^-8

Possible Answers:

0.12 M

1.5 x 10^-4 M

1.1 M

1.2 x 10^-2 M

3.0 x 10^-3 M

Correct answer:

1.5 x 10^-4 M

Explanation:

03b

$K = 9 x 10^{-8} =\frac{ [C]^2} {[A][B]} = \frac{(2x)^2}{(0.5-x)^2} \sim \frac{4x^2}{(0.25) }$

$[C] = 2 (7.5 x10^{-5} ) = 1.5 x 10^{-4}$

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Example Question #142 : Reactions And Equilibrium

What is the equilibrium concentration for A- in the following reaction if the starting concentration of HA is 0.60 M?

$HA \Leftrightarrow H^+ + A^-$ $K = 2.0 x 10^{-5}$

Possible Answers:

2.3 x 10^-3 M

2.3 x 10^-4 M

3.5 x 10^-3 M

1.2 M

1.2 x 10^-5 M

Correct answer:

3.5 x 10^-3 M

Explanation:

$K = \frac{[H^+ ][A^- ]}{[HA]} = \frac{x^2}{0.6-x} = 2.0 x 10^{-5}$

$\frac{x^2}{0.6} \sim 2.0 x 10^{-5}$

$x = 3.5 x 10^{-3}$

$[A^- ] = x = 3.5 x 10^{-3}$

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Example Question #3 : Calculating The Equilibrium Concentrations

What is the equilibrium concentration for A- in the following reaction if the starting concentration of HA is 0.30 M?

$HA \Leftrightarrow H^+ + A^-$ $K = 5.0 x 10^{-9}$

Possible Answers:

1.5 x 10^-3M

3.9 x 10^-5 M

2.1 M

2.1 x 10^-3 M

1.5 x 10^-4 M

Correct answer:

3.9 x 10^-5 M

Explanation:

$K =\frac{[H^+ ][A^- ]}{[HA]} =\frac{x^2}{0.3-x} = 2.0 x 10^{-5}$

$\frac{x^2}{0.3} \sim 5.0 x 10^{-9}$

$x = 3.87 x 10^{-5}$

$[A^- ] = x = 3.9 x 10^{-5}$

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Example Question #1 : Molecular Structure Of Acids And Bases

Put the following acids in order of their INCREASING acid strength: HI, HCl, HBr, HF.

Possible Answers:

HF, HBR, HI, HCl

HF, HCl, HBr, HI

HCl, HBr, HI, HF

HI, HBr, HCl, HF

HI, HCl, HBr, HF

Correct answer:

HF, HCl, HBr, HI

Explanation:

Larger halogen size leads to greater acidity because of weaker H-X interactions.

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Example Question #51 : Acid Base Reactions

Put the following acids in order of their INCREASING acid strength: HCl, HS, HBr, H2Se.

Possible Answers:

H2S, HCl, HBr, H2Se

H2Se, HBr, HS, HCl

H2S, H2Se, HCl, HBr

HBr, H2Se, HCl, H2S

HCl, HS, HBr, H2Se

Correct answer:

H2S, H2Se, HCl, HBr

Explanation:

Acid strength increases from left to right across a period and increases going down a group.

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AP Chemistry : Reactions and Equilibrium

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #53 : Chemical Equilibrium

Example Question #54 : Chemical Equilibrium

Example Question #55 : Chemical Equilibrium

Example Question #1 : Calculating The Equilibrium Concentrations

Example Question #56 : Chemical Equilibrium

Example Question #141 : Reactions And Equilibrium

Example Question #142 : Reactions And Equilibrium

Example Question #3 : Calculating The Equilibrium Concentrations

Example Question #1 : Molecular Structure Of Acids And Bases

Example Question #51 : Acid Base Reactions

All AP Chemistry Resources

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