Question

One mole of nitrogen and three moles of hydrogen are mixed in a 4 litre container. If 0.25 percent of nitrogen is converted to ammonia by the following reaction.
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
What will be the value of $K$ for the following equilibrium?
$\frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)\rightleftharpoons N{H}_3\left(g\right).$

• A. $1.49\times {10}^{-5}Lmo{l}^{-1}$
• B. $2.22\times {10}^{-10}Lmo{l}^{-1}$
• C. $3.86\times {10}^{-3}Lmo{l}^{-1}$
• D. None of these

Answer 1

Answer: (C)

$3.86\times {10}^{-3}Lmo{l}^{-1}$

moles $\frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)⟶N{H}_3\left(g\right)$

at $t=0$ $1$ $3$ -

$1-\frac{1}{2}\times x$ $3-\frac{3}{2}\times x$ $x$

as $\frac{1}{2}\times x=0.25\times {10}^{-2}$; $x=0.50\times {10}^{-2}$

$\Rightarrow \left[N{H}_3\right]=\frac{0.50}{4}\times {10}^{-2},\left[H_2\right]=\frac{2.9925}{4},\left[N_2\right]=\frac{0.9975}{4}$

$K_{eq}=\frac{\left[N{H}_3\right]}{\left[H_2{]}^{3/2}\right[N_2{]}^{1/2}}=\frac{\frac{0.50}{4}\times {10}^{-2}}{{\left[\frac{2.9925}{4}\right]}^{3/2}{\left[\frac{0.9975}{4}\right]}^{1/2}}$

$\Rightarrow K_{eq}=3.86\times {10}^{-3}Lmo{l}^{-1}$