Question

An air sample containing $21:79$ of $O_{2}$ and $N_{2}$ (mole ratio) is heated to $2400^\circ C$ . If the mole percent of $NO$ at equilibrium is $1.8\%,$ calculate $K_{p}$ for the reaction $N_{2}+O_{2}\rightleftharpoons2NO.$
Give your answer after multiplying by $10^{5}$ and round off up to nearest integer.

Answer 1

Given equilibrium reaction is $N_{2}+O_{2}\rightleftharpoons2NO$ .
Initial reaction,
$\underset{1}{N_{2} }+\underset{1}{O_{2} }\rightleftharpoons\underset{0}{2 NO}$
Let $aM$ of reactant will form the product.
Concentration reaction at equilibrium,
$\underset{\left(1 - a\right)}{N_{2} }+\underset{\left(1 - a\right)}{O_{2} }\rightleftharpoons\underset{2 a}{2 NO}$
An air sample containing $21:79$ of $O_{2}$ and $N_{2}$ (mole ratio).
Hence, the total moles $=0.79\left(\right.1-a\left.\right)+0.21\left(\right.1-a\left.\right)+2a$ .
At equilibrium, $NO$ is $1.8\%$
$2a=0.018Ma=0.009M$
Total moles $=1.009mole$
$K_{p}=\frac{\left(\left[NO\right]\right)^{2}}{\left[N_{2}\right] \left[O_{2}\right]}=\frac{2 \times \left(a\right)^{2}}{\left(1 - a\right)^{2}}=\frac{2 \times \left(0 . 009\right)^{2}}{\left(1 - 0 . 009\right)^{2}}K_{p}=1.6\times \left(10\right)^{- 4}.Aftermultiplyingwith\left(10\right)^{5}=16$