$2NO_{(g)} + O_{2(g)} \rightleftharpoons 2NO_{2(g)}$
What is $K$ for the reaction,
$NO_{2(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + O_{2(g)}$
$2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) ; \mathrm{K}_{2} ...(ii)$
$\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) ;$
$\mathrm{K}=\mathrm{K}_{1} \times \mathrm{K}_{2}$
therefore, For $\mathrm{NO}_{2}(\mathrm{g})=\mathrm{N}_{2} / 2(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$
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| Compound | Weight $\%$ of $P$ | Weight $\%$ of $Q$ |
| $1$ | $50$ | $50$ |
| $2$ | $44.4$ | $55.6$ |
| $3$ | $40$ | $60$ |
$(A)$ If empirical formula of compound $3$ is $P_3 Q_4$, then the empirical formula of compound $2$ is $P_3 Q_5$.
$(B)$ If empirical formula of compound $3$ is $P _3 Q _2$ and atomic weight of element $P$ is $20$ , then the atomic weight of $Q$ is $45$ .
$(C)$ If empirical formula of compound $2$ is $PQ$, then the empirical formula of the compound $1$ is $P _5 Q _4$.
$(D)$ If atomic weight of $P$ and $Q$ are $70$ and $35$ , respectively, then the empirical formula of compound $1$ is $P _2 Q$.