Question

$15$ moles of $H_2$ and $5.2$ moles of $I_2$ are mixed and allowed to attain equilibrium at ${500}^{o}C$. At equilibrium , the concentration of $HI$ is found to be $10$ moles. The equilibrium constant for the formation of $HI$ is:

• A. $50$
• B. $15$
• C. $100$
• D. $25$

Answer 1

The correct option is A $50$

$Explanation:$

Equation $H_2+I_2\to 2HI$

Given concentration $15$ $5.2$ $10$

then new

concentration: $(15-x)$ $(5.2-x)$ $2x$

when $x=$ degree of dissociation

$\because 2HI=2x=10\Rightarrow x=5$

Then,

$HI=5$

$H_2=(15-5)=10$

$I_2=5.2-5=0.5$

The equilibrium constant $=K_c=\frac{{\left[HI\right]}^2}{\left[H_2\right]\left[I_2\right]}\Rightarrow \frac{{10}^2}{10\times 0.2}\Rightarrow 50$

$\therefore K_c=50$

Hence the correct answer is option $A$