Calculating the Equilibrium Concentrations - AP Chemistry

Card 0 of 70

Question

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?

Answer

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

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

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

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

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Question

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

$M + 2L \Leftrightarrow ML2$

Answer

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

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Question

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

K = 9.0 x 10-8

Answer

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

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

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Question

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

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

Answer

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

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

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

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

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Question

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

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

Answer

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

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

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

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

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Question

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?

Answer

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

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

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

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

Compare your answer with the correct one above

Question

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

$M + 2L \Leftrightarrow ML2$

Answer

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

Compare your answer with the correct one above

Question

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

K = 9.0 x 10-8

Answer

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

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

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Question

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

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

Answer

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

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

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

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

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Question

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

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

Answer

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

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

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

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

Compare your answer with the correct one above

Question

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?

Answer

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

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

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

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

Compare your answer with the correct one above

Question

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

$M + 2L \Leftrightarrow ML2$

Answer

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

Compare your answer with the correct one above

Question

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

K = 9.0 x 10-8

Answer

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

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

Compare your answer with the correct one above

Question

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

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

Answer

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

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

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

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

Compare your answer with the correct one above

Question

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

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

Answer

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

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

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

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

Compare your answer with the correct one above

Question

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?

Answer

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

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

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

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

Compare your answer with the correct one above

Question

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

$M + 2L \Leftrightarrow ML2$

Answer

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

Compare your answer with the correct one above

Question

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

K = 9.0 x 10-8

Answer

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

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

Compare your answer with the correct one above

Question

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

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

Answer

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

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

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

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

Compare your answer with the correct one above

Question

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

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

Answer

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

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

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

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

Compare your answer with the correct one above

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Calculating the Equilibrium Concentrations - AP Chemistry

Consider formation of Nitrogen monoxide: with . If the initial concentration of N2 was 0.085M and O2 was 0.038 M, what is the concentration of nitrogen monoxide at equilibrium?

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

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

Find the equilibrium concentrations for C in the following chemical reaction: A + B -> 2C K = 9.0 x 10-8

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

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

Consider formation of Nitrogen monoxide: with . If the initial concentration of N2 was 0.085M and O2 was 0.038 M, what is the concentration of nitrogen monoxide at equilibrium?

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

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

Find the equilibrium concentrations for C in the following chemical reaction: A + B -> 2C K = 9.0 x 10-8

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

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

Consider formation of Nitrogen monoxide: with . If the initial concentration of N2 was 0.085M and O2 was 0.038 M, what is the concentration of nitrogen monoxide at equilibrium?

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

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

Find the equilibrium concentrations for C in the following chemical reaction: A + B -> 2C K = 9.0 x 10-8

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

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

Consider formation of Nitrogen monoxide: with . If the initial concentration of N2 was 0.085M and O2 was 0.038 M, what is the concentration of nitrogen monoxide at equilibrium?

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

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

Find the equilibrium concentrations for C in the following chemical reaction: A + B -> 2C K = 9.0 x 10-8

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

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

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?

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

$M + 2L \Leftrightarrow ML2$

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

K = 9.0 x 10-8

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

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

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

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

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?

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

$M + 2L \Leftrightarrow ML2$

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

K = 9.0 x 10-8

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

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

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

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

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?

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

$M + 2L \Leftrightarrow ML2$

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

K = 9.0 x 10-8

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

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

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

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

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?

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

$M + 2L \Leftrightarrow ML2$

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

K = 9.0 x 10-8

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

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

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

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