Question

Q) A vessel contains N2O4 and NO2 in 2:3 molar ratio at 10 atm under equilibrium. Now Kp for

N2O4 ----2NO2 is

( ans: 9 atm)

Answer 1

N2O4  - - - - - - - > 2 NO2

At equilibrium ,

x                                1 - x

Thus , x = 2 / 5

1 - x = 3 / 5

P a = x P

P N2O4 = 2 / 5 x 10 = 4 atm

P NO2 = 3 / 5 x 10 = 6 atm

Kp = P NO2 ^ 2 / P N2O4 = 36 / 4

Kp = 9 atm

Answer 2

Answer: The value of $K_p$ for above equation is 9 atm

Explanation:

We are given:

Mole ratio of $N_2O_4\text{ and }NO_2$ = 2 : 3

So, moles of $N_2O_4$ = 2 moles

Moles of $NO_2$ = 3 moles

Mole fraction of $N_2O_4=\frac{n_{N_2O_4}}{n_{N_2O_4}+n_{NO_2}}=\frac{2}{5}$

Mole fraction of $NO_2=\frac{n_{NO_2}}{n_{N_2O_4}+n_{NO_2}}=\frac{3}{5}$

Total pressure of the system = 10 atm

The partial pressure of a gas is given by Raoult's law, which is:

$p_A=p_T\times \chi_A$

• For $N_2O_4$ :

Using above equation:

$p_{N_2O_4}=\frac{2}{5}\times 10=4atm$

• For $NO_2$ :

Using above equation:

$p_{NO_2}=\frac{3}{5}\times 10=6atm$

For the given chemical equation:

$N_2O_4\rightleftharpoons 2NO_2$

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{NO_2}^2}{p_{N_2O_4}}\\\\K_p=\frac{6\times 6}{4}\\\\K_p=9atm$

Hence, the value of $K_p$ for above equation is 9 atm