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. $\left[I_{2}\right]<\left[H_{2}\right], \left[I_{2}\right] = \left[HI\right]$
• D. $\left[I_{2}\right]>\left[H_{2}\right], \left[I_{2}\right] = \left[HI\right]$

Answer 1

Answer: (B)

For the given reaction, $K = \frac{\left[HI\right]^{2}}{\left[H_{2}\right]\left[I_{2}\right]}$
As 1 mole of H$_2$ reacts with 1 mole of I$_2$, even at equilibrium, $[H_2] = [I_2]$
Hence, $K = \frac{\left[HI\right]^{2}}{\left[I_{2}\right]^{2}}\quad$ or$\quad\sqrt{K}=\frac{\left[HI\right]}{\left[I_{2}\right]}=\sqrt{47.6}$
i.e., $\left[HI\right]>\left[I_{2}\right]$