Question

Reaction between nitrogen and oxygen takes place as follows:
${2N}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons {2N}_{2}O\left(g\right)$
If a mixture of $0.482$ of $N_2$ and $0.933$ mol of $N_2$ is placed in a reaction in a reaction vessel of volume $10L$ and allowed to form $N_{2}O$ at a temperature for which $K_c=2.0\times {10}^{-37}$. Determine the composition of the equilibrium mixture.

Answer 1

$2N_2+O_2\rightleftarrows 2N_{2}O$

Initial $2\times 0.482$ $0.933$ $0$

Equilibrium

$.964-2x$ $0.933-x$ $2x$

$K_c=2\times {10}^{-37}$

Volume $=10lit$ (Given)

$K_c=\frac{[N_{2}O{]}^2}{\left[O_2\right][N_2{]}^2}$

$2\times {10}^{-37}=\frac{4x^2}{(0.933-x)(.964-2x{)}^2}$

$\because x<<<1$

$0.933-x=0.933$

$4x^2=2\times {10}^{-37}\times 0.933\times (0.964{)}^2$

$x=0.065\times {10}^{-18}mole$

Composition of $N_2=.482-0.065\times {10}^{-18}$

$=0.482mole$

$O_2=0.933-0.065\times {10}^{-18}$

$=0.0933mole$

$N_{2}O=0.065\times {10}^{-18}$