Question

The following concentrations were obtained for the formation of $N{H}_3$ from $N_2\text{and}{N}_2$ at equilibrium at $500K$. $\left[N_2\right]=1.5\times {10}^{-2}M$. $\left[H_2\right]=3.0\times {10}^{-2}M$ and $\left[N{H}_3\right]=1.2\times {10}^{-2}M$. Calculate equilibrium constant.

Answer 1

Step 1: Equilibrium constant
For a general reversible reaction $aA+bB\rightleftharpoons cC+dD$, the concentrations in an equilibrium mixture are related by the following equilibrium equation.

$K_c=\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B{]}^b}$

Step 2: Given reaction
Given reaction can be written as-
$N_2\left(g\right)+H_2\left(g\right)\rightleftharpoons N{H}_3\left(g\right)$

The balanced reaction can be written as-
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$

Step 3: Value of $K_c$

Therefore, for the given reaction,
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$, equilibrium constant can be written as-
$K_c=\frac{\left[N{H}_3\right(g){]}^2}{\left[N_2\right(g\left)\right]\left[H_2\right(g){]}^3}$[]

$=\frac{(1.2\times {10}^{-2}{)}^2}{(1.5\times {10}^{-2})(3.0\times {10}^{-2}{)}^3}$

$=0.106\times {10}^4=1.06\times {10}^3$

Final answer:
The value of equilibrium constant is: $1.06\times {10}^3$.