Question

How many moles of H₂ and HI will be present at equilbrium if 0.1570 mol HI are placed into a 1-L flask and allowed to react at a temperature where for the following reaction takes place:

2HI(g)  H₂(g) + I₂(g)K = 10.8

Answer 1

Concentration of HI initially$=\frac{0.1570}{1}=0.1570M$

$2HI(g) = H_2(g) + I_2(g)$ $K = 10.8$

$K=\frac{(H_2)(1_2)}{(HI)^2}$ =$\frac{(x)(x)}{(0.1570-2x)^2}=10.8$

$K=[\frac{x}{0.1570+2x}]^2=10.8$

$K=[\frac{x}{0.1570-2x}]=3.29$

$0.1570-2x=3.29x$

$x= 0.03$ hence the equlibrium

H₂ =0.03