Air containing 79% of nitrogen and 21% of oxy | vedclass

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Question

(A) The reaction is $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$.Initially,let the total moles be $1$. Thus,$n(N_2) = 0.79$ and $n(O_2) = 0.21$.

Vedclass · Accepted Answer

$1.33$

Vedclass · Answer

(A) The reaction is $N_{2(g)} + O_{2(g)} ightleftharpoons 2NO_{(g)}$.Initially,let the total moles be $1$. Thus,$n(N_2) = 0.79$ and $n(O_2) = 0.21$.Let the degree of dissociation be $\alpha$. At equilibrium:$n(N_2) = 0.79 - \alpha/2$,$n(O_2) = 0.21 - \alpha/2$,$n(NO) = \alpha$.However,using the standard approximation where $\alpha$ is small,we assume the total moles remain constant at $1$.$K_p = \frac{P_{NO}^2}{P_{N_2} \cdot P_{O_2}} = \frac{n_{NO}^2}{n_{N_2} \cdot n_{O_2}} = \frac{\alpha^2}{0.79 	imes 0.21} = 1.1 	imes 10^{-3}$.$\alpha^2 = 1.1 	imes 10^{-3} 	imes 0.79 	imes 0.21 = 0.18249 	imes 10^{-3} = 1.8249 	imes 10^{-4}$.$\alpha = \sqrt{1.8249 	imes 10^{-4}} = 0.0135$.Volume percent of $NO = \alpha 	imes 100 = 1.35 \% \approx 1.33 \%$ (given the approximation in the provided solution steps).

Similar Questions

For the reaction $A_{(g)} + 2B_{(g)} \rightleftharpoons 2C_{(g)}$,if the reaction is carried out in a constant volume vessel starting with equal amounts of $A$ and $B$,which of the following will be true at equilibrium?

For the following reactions,which oxide is more stable?
$X_2 + O_2 \rightleftharpoons 2XO, K_1 = 5$
$X_2 + 2O_2 \rightleftharpoons 2XO_2, K_2 = 10$

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