Question

$18.4g{N}_2{O}_4$ was placed in $1L$ vessel at $400K$ and allowed to attain the following equilibrium $N_2{O}_4(g)\rightleftharpoons 2N{O}_2(g)$
If the total pressure at equilibrium was $10.64$ bar, approximate $K_p$ is $(R=0.083L$ bar $K^{-1}mo{l}^{-1})$ (Assume $N_2{O}_4,N{O}_2$ as ideal gases)

• A. 57.20
• B. 24.24
• C. 14.30
• D. 6.64

Answer 1

Answer: (B)

From ideal gas equation,

$pV=nRT$

$p_{N_2{O}_4}\times 1=\frac{18.4}{92}\times 0.083\times 400$

$p_{N_2{O}_4}=6.64$ bar

For the equilibrium reaction,

	$N_2{O}_4(g)$	$\rightleftharpoons$	$2N{O}_2(g)$
Initial pressure	P		0
At equilibrium	$P-p_i$		$2p_i$

$p_T=p-p_i+2p_i$

$p_T=p+p_i$

$\Rightarrow p_i=p_T-p$

$p_i=10.64-6.64$

$=4.00$

At equilibrium,

$\therefore p_{N_2{O}_4}=p-p_i=6.64-4$

$=2.64$ bar

$p_{N{O}_2}=2p_i=2\times 4$

$=8$ bar

$\therefore K_p=\frac{{\left[p_{N{O}_2}\right]}^2}{\left[p_{N_2{O}_4}\right]}$

$=\frac{[8{]}^2}{2.64}=24.29$