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AP Chemistry : Calculating the Equilibrium Concentrations

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Example Question #1 : Calculating The Equilibrium Concentrations

Consider formation of Nitrogen monoxide: $N_2 (g) + O_2 (g) \Leftrightarrow 2 NO (g)$ $\displaystyle N_2 (g) + O_2 (g) \Leftrightarrow 2 NO (g)$ with $K = 4.2 x 10^{-8}$ $\displaystyle K = 4.2 x 10^{-8}$. 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:

3.2 x 10^-8 M

1.1 x 10^-3 M

1.2 x 10^-5 M

1.2 x 10^-6 M

1.1 x 10^-4M

Correct answer:

1.2 x 10^-5 M

Explanation:

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

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

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

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

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Example Question #2 : Calculating The Equilibrium Concentrations

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

$M + 2L \Leftrightarrow ML2$ $\displaystyle M + 2L \Leftrightarrow ML2$

Possible Answers:

-x, -x, +x

+x, +2x, +x

+x, +x, +x

+x, +4x, -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 #1 : Calculating The Equilibrium Concentrations

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

K = 9.0 x 10^-8

Possible Answers:

3.0 x 10^-3 M

1.5 x 10^-4 M

1.1 M

0.12 M

1.2 x 10^-2 M

Correct answer:

1.5 x 10^-4 M

Explanation:

03b

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

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

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Example Question #4 : Calculating The Equilibrium Concentrations

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

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

Possible Answers:

2.3 x 10^-4 M

3.5 x 10^-3 M

2.3 x 10^-3 M

1.2 x 10^-5 M

1.2 M

Correct answer:

3.5 x 10^-3 M

Explanation:

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

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

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

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

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Example Question #2 : 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?

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

Possible Answers:

2.1 M

1.5 x 10^-3M

1.5 x 10^-4 M

3.9 x 10^-5 M

2.1 x 10^-3 M

Correct answer:

3.9 x 10^-5 M

Explanation:

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

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

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

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

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AP Chemistry : Calculating the Equilibrium Concentrations

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Example Question #1 : Calculating The Equilibrium Concentrations

Example Question #2 : Calculating The Equilibrium Concentrations

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