Question

Density of equilibrium mixture of $N_{2}O$ and $N{O}_2$ at 1 atm and 384 K is $1.84g/d{m}^3$. Equilibrium constant of the following reaction is:

$N_2{O}_4\rightleftharpoons 2N{O}_2$

• A. 1.98 atm
• B. 2.09 atm
• C. 2.36 atm
• D. 1.48 atm

Answer 1

The correct option is B 2.09 atm

The molecular weight

$M=\frac{\rho RT}{P}=\frac{1.84\times 0.08206\times 384}{1}=58$ g/mol.
Now,
$N_2{O}_4\rightleftharpoons 2N{O}_2$
Let, $x$ be the mole fraction of $N_2{O}_4$. The mole fraction of $N{O}_2$ will be $1-x$
The molecular weights of $N{O}_2$ and $N_2{O}_4$ are 46 g/mol and 92 g/mol respectively.
The average molecular weight of the equilibrium mixture is 58.
$58=92x+46(1-x)$
$58=92x+46-46x$
$58=46+46x$
$12=46x$
$x=0.26$

$1-x=1-0.26=0.74$
The partial pressure is the product of the mole fraction and total pressure (11 atm).
The equilibrium constant $K_p=\frac{P_{N{O}_2}^2}{P_{N_2{O}_4}}=\frac{(1\times 0.74{)}^2}{1\times 0.26}=2.09$ atm