Question

Consider the reaction below that has a Keq of 3.43 ⋅ 10 − 5 . 1.0...

Consider the reaction below that has a Keq of 3.43 ⋅ 10 − 5 . 1.0 mol of A and 7.4 mol of B are added to a 8 L container. Calculate the equilibrium concentration of C. Enter your answer in scientific notation using 2 significant figures.

2 A + 3 B ↔ 1 C

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Answer #1

The equilibrium constant K_{eq} = \frac{[C]}{[A]^2 \times [B]^3}=3.43 \times 10^{-5}

First calculate initial concentrations

\\ \ \\ \ [A] = \frac{1.0 \ mol}{8 \ L} =0.125 \ M \\ \ \\ \ [B]= \frac{7.4 \ mol}{8 \ L} =0.925 \ M

Prepare an ICE table.

A B C
Initial concentration (M) 0.125 0.925 0.00
Change in concentration (M) -2x -3x +x
Equilibrium concentration (M) 0.125-2x 0.925-3x x

Substitute values in equilibrium constant expression.

K_{eq} = \frac{[C]}{[A]^2 \times [B]^3}=3.43 \times 10^{-5}

K_{eq} = \frac{x}{(0.125-2x)^2 \times (0.925-3x)^3}=3.43 \times 10^{-5}

Since, the value of the equilibrium constant is very small, approximate 0.125-2x to 0.125 and 0.925-3x to 0.925

K_{eq} = \frac{x}{(0.125)^2 \times (0.925)^3}=3.43 \times 10^{-5}

x=3.43 \times 10^{-5} \times (0.125)^2 \times (0.925)^3

x=4.24 \times 10^{-7}

The equilibrium concentration of C is x=4.2 \times 10^{-7} \ M

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