Calculating the Equilibrium Concentrations - AP Chemistry

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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?

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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?

$M + 2L \Leftrightarrow ML2$

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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

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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}$$

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Answer

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

$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}$$

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Answer

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

$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?

Tap to reveal answer

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?

$M + 2L \Leftrightarrow ML2$

Tap to reveal answer

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

Tap to reveal answer

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}$$

Tap to reveal answer

Answer

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

$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}$$

Tap to reveal answer

Answer

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

$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?

Tap to reveal answer

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?

$M + 2L \Leftrightarrow ML2$

Tap to reveal answer

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

Tap to reveal answer

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}$$

Tap to reveal answer

Answer

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

$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}$$

Tap to reveal answer

Answer

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

$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?

Tap to reveal answer

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?

$M + 2L \Leftrightarrow ML2$

Tap to reveal answer

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

Tap to reveal answer

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}$$

Tap to reveal answer

Answer

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

$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}$$

Tap to reveal answer

Answer

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

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

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

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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}$$