Question

A mixture of 0.3 mole of $H_{2}$ and 0.3 mole of $I_{2}$ is allowed to react in a 10 litre evacuated flask at $500^{o}C$. The reaction is H₂ + I₂ ⇌ 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

Answer 2

$K_{c} = \frac{\lbrack HI\rbrack^{2}}{\lbrack H_{2}\rbrack\lbrack I_{2}\rbrack}$; $\mathbf{64 =}\frac{\mathbf{x}^{\mathbf{2}}}{\mathbf{0.03}\mathbf{\times}\mathbf{0.03}}$

$\mathbf{x}^{\mathbf{2}}\mathbf{= 64}\mathbf{\times}\mathbf{9}\mathbf{\times}\mathbf{1}\mathbf{0}^{\mathbf{-}\mathbf{4}}$; $\mathbf{x = 8}\mathbf{\times}\mathbf{3}\mathbf{\times}\mathbf{1}\mathbf{0}^{\mathbf{-}\mathbf{2}}\mathbf{= 0.24}$

$x$ is the amount of $HI$ at equilibrium. Amount of $I_{2}$ at

equilibrium will be $0.30 - 0.24 = 0.06$mole