Question

Evaluate $K_p$ for the reaction : $H_2+I_2\rightleftharpoons 2HI$. $2$ moles each of $H_2$ and $I_2$ are taken initially equilibrium moles of $HI$ are $2$ :

• A. $2.5$
• B. $4$
• C. $0.25$
• D. $1.0$

Answer 1

The correct option is B $4$

Reaction;

$H_2+I_2\to 2HI$

$K_C=\frac{(HI{)}^2}{\left(H_2\right)\left(I_2\right)}$

Initial moles of $H_2$ and $I_2$ =2

At equilibrium ,moles of $H_2$ and $I_2$ =2-x

At equilibrium moles of HI=2

From reaction x mole of $H_2$ and $I_2$ will give 2x mole of 2HI

Then, $2x=2$

$x=1$

$K_C=\frac{(2{)}^2}{\left(1\right)\left(1\right)}$

=$4$