Question

Calculate the concentration of $H_2,I_2$ and $HI$ at equilibrium if $0.0400$ mole of $HI$ is placed in a $1.00L$ flask at ${430}^{o}C$. $K_c$ for the reaction:
$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$ is $54.4$ at this temperature.

• A. $\left[H_2\right]\left[M\right]=0.0200\left[I_2\right]\left[M\right]=0.0200\left[HI\right]\left[M\right]=0.0200$
• B. $\left[H_2\right]\left[M\right]=0.00427\left[I_2\right]\left[M\right]=0.00427\left[HI\right]\left[M\right]=0.0315$
• C. $\left[H_2\right]\left[M\right]=0.0315\left[I_2\right]\left[M\right]=0.0315\left[HI\right]\left[M\right]=0.00850$
• D. $\left[H_2\right]\left[M\right]=0.000478\left[I_2\right]\left[M\right]=0.00478\left[HI\right]\left[M\right]=0.0352$

Answer 1

The correct option is B $\left[H_2\right]\left[M\right]=0.00427\left[I_2\right]\left[M\right]=0.00427\left[HI\right]\left[M\right]=0.0315$

Since total volume is 1.00 L, the number of moles is equal to molar concentration.

	$H_2$	$I_2$	$HI$
Initial concentration (M)	0	0	0.0400
Change in concentration (M)	$+x$	$+x$	$-2x$
Equilibrium concentration (M)	$x$	$x$	$0.0400-2x$

The equilibrium constant $K_c=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$

$54.4=\frac{(0.0400-2x{)}^2}{x\times x}$

$54.4=(\frac{(0.0400-2x)}{x}{)}^2$

$\sqrt{54.4}=\left(\frac{(0.0400-2x)}{x}\right)$

$7.38=\left(\frac{(0.0400-2x)}{x}\right)$

$7.38x=0.0400-2x$

$9.38x=0.0400$

$x=0.00427$

The equilibrium concentrations are

$\left[H_2\right]=\left[I_2\right]=x=0.00427M$

$\left[HI\right]=0.0400-2x=0.0400-2\left(0.00427\right)=0.0315M$