Question

1 mole of $H_2,2$ moles of $I_2$ and 3 moles of $HI$ were taken in $1$ litre flask. The equilibrium constant of the reaction $H_{2\left(g\right)}+I_{2\left(g\right)}\rightleftharpoons 2H{I}_{\left(g\right)}$ is $50$ at ${330}^{o}C.$ The concentration of $HI$ at equilibrium is: (moles.lit${}^{-1}$ )

• A. 0.3
• B. 1.3
• C. 4.4
• D. 2.7

Answer 1

Answer: (C)

4.4

Equation for the reaction:

$H_2\left(g\right)+I_2\left(g\right)\to 2HI\left(g\right)$

Initial moles: $1$ $2$ $3$

Moles at equilibrium $1-x$ $2-x$ $3+2x$

Formula for the equilibrium constant:

$K_c=\frac{{\left[HI\right]}^2}{\left[H_2\right]\left[I_2\right]}=50$

$50=\frac{{\left(3+2x\right)}^2}{\left(1-x\right)\left(2-x\right)}$

$\Rightarrow x\approx 0.7$

Substitute value of $x$ in $3+2x$ ;

We get concentration of HI as $3+2\times 0.7=4.4M\left(mol/lit\right)$

Therefore option C is correct.