A reaction mixture containing $\mathrm{H}_{2}, \mathrm{~N}_{2}$ and $\mathrm{NH}_{3}$ has partial pressures…
- Forward
- Backward
- No net reaction
- Direction of reaction cannot be predicted
Solution
\mathrm{O}_{\mathrm{p}}=\frac{\left(\mathrm{P}_{\mathrm{NH}_{3}}ight)^{2}}{\mathrm{P}_{\mathrm{N}_{2}} \times\left(\mathrm{P}_{\mathrm{H}_{2}}ight)^{3}}=\frac{(3)^{2}}{(1)(2)^{3}}=\frac{9}{8} \mathrm{~atm}^{-2}
$$
$$
=1.125 \mathrm{~atm}^{-2}
$$
Since value of $\mathrm{Q}_{\mathrm{p}}$ is larger than $\mathrm{K}_{\mathrm{p}}$ $\left(4.28 \times 10^{-5} \mathrm{~atm}^{-2}ight)$, it indicates net reaction will proceed in backward direction.
Asked in: JEE-TOPICTESTS-CHEMISTRY