Question

13.8g of $N_2{O}_4$ was placed in a 1 L reaction vessel at 400 K and allowed to attain equilibrium $N_2{O}_4\rightleftharpoons 2N{O}_2$
The total pressure at equilibrium was found to be 9.15bar. Calculate $K_P,K_C$ and partial pressure at equilibrium.

Answer 1

We know that, $PV=nRT$

molar mass of $N_2{O}_4=92g$

number of moles $=\frac{13.8}{92}=0.15$ of the gas

$R=0.083$ bar $L$ mol

$T=400K$

So, $PV=nR{T}_p\times 1L$

$=0.15\times 0.083\times 400=4.98$ bar

If $N_2{O}_4$ ${2NO}_2$ initial pressure : $4.98$ bar $O$ at eqn.

: $(4.98-\lambda )$ bar $2x$ bar

Hence

$P_{Total}=P_{N_2{O}_4}+P_{{NO}_2}$

$9.15=(4.98-x)+2x9.5=4.98+xx$

$=9.15-4.98=4.17$ bar

$K_p=\frac{\left(8.34\right)\times 2}{0.81}$

$\Rightarrow K_p=85.87$

Now,

$K_p=K_c\left(RT\right)\Delta n$

$\Rightarrow 85.87=K_c(0.083\times 400)$

$\Rightarrow K_c=\frac{85.87}{0.083\times 400}$

$\Rightarrow K_c=2.586$