Answer to Question #286962 in General Chemistry for Amahle

Question

The equilibrium concentrations of in a gas mixture at a particular temperature are 0.13 mol. dm-3 H2, 0.70 mol. dm-3 I2, and 2.1 mol. dm-3 HI. What equilibrium concentrations are obtained when 0.200 mol of HI is injected into an empty 500 cm3 container?

Answer 1

$H_2+I_2\rightleftharpoons2HI$

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

$K_c=\frac{(2.1)^2}{0.13×0.7}=48.46$

$[HI]=\frac{0.2}{0.5}=0.4m$

$\begin{matrix} & &H_2&&I_2&&2HI \\ I&&0&&0&&0.4 \\ C&&x&&x&&-2x\\ E&&x&&x&&0.4-2x \end{matrix}$

$48.46=\frac{(0.4-2x)^2}{(x)(x)}$

$6.961=\frac{0.4-2x}{x}$

$x=0. 04464$

$[HI]=0.4-2(0.04464)$

$=0.31072$

$[H_2]=[I_2]=0.04464$