Question

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

Answer 1

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

First calculate initial concentrations

1.0 mol A 0.125 M 8 L 7.4 mol 0.925 M 8 L

Prepare an ICE table.

Initial concentration (M)
Change in concentration (M)
Equilibrium concentration (M)

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 $x=4.2 \times 10^{-7} \ M$