Reaction between $\mathrm{N}_2$ and $\mathrm{O}_2$ takes place as follows : $2 \mathrm{~N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}(\mathrm{g})$ If a mixture of $0.482 \mathrm{~mol} \mathrm{~N}_2$ and $0.933 \mathrm{~mol} \mathrm{of} \mathrm{O}_2$ is placed in a $10 \mathrm{~L}$ reaction vessel and allowed to form $\mathrm{N}_2 \mathrm{O}$ at a temperature for which $\mathrm{K}_{\mathrm{c}}=2.0 \times 10^{-37}$. Determine the composition of equilibrium mixture.

Question

Reaction between $\mathrm{N}_2$ and $\mathrm{O}_2$ takes place as follows : $2 \mathrm{~N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}(\mathrm{g})$ If a mixture of $0.482 \mathrm{~mol} \mathrm{~N}_2$ and $0.933 \mathrm{~mol} \mathrm{of} \mathrm{O}_2$ is placed in a $10 \mathrm{~L}$ reaction vessel and allowed to form $\mathrm{N}_2 \mathrm{O}$ at a temperature for which $\mathrm{K}_{\mathrm{c}}=2.0 \times 10^{-37}$. Determine the composition of equilibrium mixture.,

MARKS App · Accepted Answer

As $\mathrm{K}_{\mathrm{c}}=2.0 \times 10^{-37}$ is very small, this means that the amount of $\mathrm{N}_2$ and $\mathrm{O}_2$ reacted $(x)$ is very very small and can be neglected. Hence, at equilibrium, we have $\left[\mathrm{N}_2\right]=0.0482 \mathrm{~mol} \mathrm{~L}^{-1}$, $\left[\mathrm{O}_2\right]=0.0933 \mathrm{~mol} \mathrm{~L}^{-1},\left[\mathrm{~N}_2 \mathrm{O}\right]=0.1 x$ $\mathrm{K}_{\mathrm{c}}=\frac{(0.1 x)^2}{(0.0482)^2(0.0933)}=2.0 \times 10^{-37}$ On solving, this gives $x=6.57 \times 10^{-20}$ $\left[\mathrm{N}_2 \mathrm{O}\right]=0.1 x=6.57 \times 10^{-21} \mathrm{~mol} \mathrm{~L}^{-1}$

Search any question & find its solution

Looking for more such questions to practice?