Question

For a reaction, $2HI\rightleftharpoons H_2+I_2$ at equilibrium, $7.8$ g, $203.2$ g and $1638.4$ g of $H_2$, $I_2$ and $HI$ respectively were found. Calculate $K_C$.

• A. $K_C=2.1$
• B. $K_C=0.019$
• C. $K_C=0.005$
• D. $K_C=0.27$

Answer 1

The correct option is B $K_C=0.019$

$2HI\rightleftharpoons H_2+I_2$

$\frac{1638.4}{128}$ $\frac{7.8}{2}$ $\frac{203.2}{254}$ Moles at equilibrium

$=12.8$ $=3.9$ $=0.8$

Let volume of container be $V$ L.

$\therefore K_C=\frac{\left[H_2\right]\left[I_2\right]}{{\left[HI\right]}^2}=\frac{3.9\times 0.8}{V\times V\times [\frac{12.8}{V}{]}^2}=0.019$

$\therefore K_C=0.019$