Question

One mole of nitrogen is mixed with three moles of hydrogen in a four litre container. If $0.25$ percent of nitrogen is converted to ammonia by the following reaction
$N _{2}(g)+3 H _{2}(g) \rightleftharpoons 2 NH _{3}(g)$
then calculate the equilibrium constant, $K_{c}$ in concentration units. What will be the value of $K_{c}$ for the following equilibrium
$\frac{1}{2} N_{2}(g)+\frac{3}{2} H_{2}(g) \rightleftharpoons 2 N H_{3}(g)$

Answer 1

${\left[N_{2}\right]=\frac{0.75}{4},\left[H_{2}\right]=\frac{2.25}{4},\left[N H_{3}\right]=\frac{0.50}{4} }$
$K_{c}=\frac{\left[N H_{3}\right]^{2}}{\left[N_{2}\right]\left[H_{2}\right]^{3}}=\frac{(0.50)^{2}}{(0.75)(2.25)^{3}} \times 16$
$=0.468 L^{2} m o l^{-2}$
Also for: $\frac{1}{2} N_{2}+\frac{3}{2} H_{2} \rightleftharpoons N H_{3}$
$K_{c}^{\prime}=\sqrt{K_{c}}=0.68$