Question

A mixture of $0.3$ mole of $H_2$ and $0.3$ mole of $I_2$ is allowed to react in a $10$ lit. evacuated flask at ${500}^{0}C$. The reaction is $H_2+I_2\rightleftharpoons 2HI$, the $K_c$ is found to be $64$. The amount of unreacted $I_2$ at equilibrium is:

• A. $0.15$ mole
• B. $0.06$ mole
• C. $0.03$ mole
• D. $0.2$ mole

Answer 1

Answer: (B)

$0.06$ mole

$H_2+I_2\to 2HI$

Initial $0.3$ $0.3$ --

At Equilibrium $0.3-x$ $0.3-x$ $2x$

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

$64=\frac{{\left[\frac{2x}{10}\right]}^2}{\left[\frac{0.3-x}{10}\right]\left[\frac{0.3-x}{10}\right]}$

$60x^2-38.4x+5.76=0$

$x=0.24$

Hence,moles of $I_2$ unreacted $=0.3-0.24=0.06$ mole.

Therefore, option $B$ is correct.