Question

For the reaction, $H_2+I_2\rightleftharpoons 2HI,K=47.6$. If the initial number of moles of each reactant and product is 1 mole, then at equilibrium:

• A. $\left[I_2\right]=\left[H_2\right],\left[I_2\right]>\left[HI\right]$
• B. $\left[I_2\right]=\left[H_2\right],\left[I_2\right]<\left[HI\right]$
• C. $[I-2]<\left[H_2\right],\left[I_2\right]=\left[HI\right]$
• D. $[I-2]>\left[H_2\right],\left[I_2\right]=\left[HI\right]$

Answer 1

The correct option is B $\left[I_2\right]=\left[H_2\right],\left[I_2\right]<\left[HI\right]$

For the given reaction , $K=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$

As one mole of $H_2$ reacts with 1 mole of $I_2$ , even at equilibrium , $\left[H_2\right]=\left[I_2\right]$

Hence, $K=\frac{[HI{]}^2}{[I_2{]}^2}$ or $\sqrt{K}=\frac{\left[HI\right]}{\left[I_2\right]}=\sqrt{47.6}$

i.e , $\left[HI\right]>\left[I_2\right]$

Hence, option B is correct.