Equal before law!!!!

Chemistry Level 3

$HNO_3$ on dehydration produces two products A and B. B is a Bronsted acid as well as Bronsted base.

Now, A on thermal decomposition produces 2 products C and D. Now, D is not a compound.

Compound E exists in equilibrium with C only.

If the vapour density of E at certain temperature is 30.66 and total pressure is 3 atm , Then what is the value of $K_p$ in atm?

1 solution

Aryan Sanghi
May 30, 2019

$HNO_3$ on dehydration produces $N_2O_5$ and $H_2O$ . Now, as B is $H_2O$ , therefore A is $N_2O_5$ .

Now, A on thermal decomposition produces $NO_2$ and $O_2$ . As $O_2$ is not a compound, it is D. Therefore, C is $NO_2$ .

As $NO_2$ exists in equilibrium with $N_2O_4$ , E is $N_2O_4$ .

Now, Degree of dissociation $\alpha$ = $\frac{Theoretical Vapour density(D)-Experimental Vapour density (d)}{Experimental Vapour density (d)}$

$\alpha$ = $\frac{46-30.66}{30.66}$ =0.5

Now $K_p$ = $\frac{[p_{NO_2}]^2}{[p_{N_2O_4}]}$

$K_p$ = $\frac{4\alpha^2}{1-\alpha^2}$ × $P_{Total}$

$K_p$ =4atm