Question

If a mixture of 0.482 mol $N_2$ and 0.933 mol of $O_2$ is placed in a 10 L reaction vessel and allowed to form $N_{2}O$ at a temperature for which $K_C=2.0\times {10}^{-37}$. The composition of equilibrium mixture are :

• A. $N_2=0.482mol/L,O_2=0.0933mol/L,N_{2}O=6.6\times {10}^{-21}mol/L$
• B. $N_2=0.482mol/L,O_2=0.933mol/L,N_{2}O=6.6\times {10}^{-21}mol/L$
• C. $N_2=0.0482mol/L,O_2=0.0933mol/L,N_{2}O=6.6\times {10}^{-21}mol/L$
• D. $N_2=0.482mol/L,O_2=0.0931mol/L,N_{2}O=6.6\times {10}^{-21}mol/L$

Answer 1

Answer: (C)

$N_2=0.0482mol/L,O_2=0.0933mol/L,N_{2}O=6.6\times {10}^{-21}mol/L$

The equilibrium reaction is $2N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2N_{2}O\left(g\right)$.
The initial number of moles of $N_2,O_2$ and $N_{2}O$ are $0.482$ mol, $0.933$ mol and $0$ mol respectively.
The volume is $10$ L, The equilibrium concentration of $N_2,O_2$ and $N_{2}O$ are $(0.0482-2x)M,(0.0933-x)M$ and $2xM$ respectively.
The expression for the equilibrium constant is $K_c=2\times {10}^{-37}=\frac{(2x{)}^2}{(0.0482-2x{)}^2\times (0.0933-x)}$
The value of $4K_c$ is small, hence the value of x is very small.

$0.0482-2x\simeq 0.0482$ and $0.0933-x\simeq 0.0933$.
The expression for the equilibrium constant becomes $K_c=2\times {10}^{-37}=\frac{4x^2}{{0.0482}^2\times 0.0933}$.
Thus, $x=3.3\times {10}^{-21}$.
Hence, $\therefore \left[N_{2}O\right]=2x=6.6\times {10}^{-21}$.
The equilibrium concentrations are $\left[N_2\right]\simeq 0.0482M;\left[O_2\right]=0.0993M$.