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 $N_2+O_2\rightleftharpoons 2NO.$
Give your answer after multiplying by $10^5$ and round off up to nearest integer.

Answer 1

Answer: 16

Given equilibrium reaction is $N_2+O_2\rightleftharpoons 2NO$ .
Initial reaction,
$\underset{1}{N_2}+\underset{1}{O_2}\rightleftharpoons \underset{0}{2NO}$
Let $aM$ of reactant will form the product.
Concentration reaction at equilibrium,
$\underset{\left(1-a\right)}{N_2}+\underset{\left(1-a\right)}{O_2}\rightleftharpoons \underset{2a}{2NO}$
An air sample containing $21:79$ of $O_2$ and $N_2$ (mole ratio).
Hence, the total moles $=0.79(1-a)+0.21(1-a)+2a$ .
At equilibrium, $NO$ is $1.8\%$
$2a=0.018Ma=0.009M$
Total moles $=1.009mole$
$K_p=\frac{{\left(\left[NO\right]\right)}^2}{\left[N_2\right]\left[O_2\right]}=\frac{2\times {\left(a\right)}^2}{{\left(1-a\right)}^2}=\frac{2\times {\left(0.009\right)}^2}{{\left(1-0.009\right)}^2}{K}_p=1.6\times {\left(10\right)}^{-4}.Aftermultiplyingwith{\left(10\right)}^5=16$