Question

For the reaction equilibrium, 2NOBr (g) 2NO (g) + Br2 (g), If PBr2 = P/9 at equilibrium and P is total pressure. The ratio Kp/P is equal to:

Answer 1

equilibrium:
Kp= partial pressure of NO ² x Partial pressure of Br₂/ Partial pressure of NOBr
= (2 x P/ 9)² x P/9 / (2 x P/3)²
= (4 x P³/729) / (4 x P²/9)
= P/81
so Kp/P will be = P/81/P= 1/81

Answer 2

$\mathrm{The \ ratio \ of \ {K}_{\mathrm{p}} : \mathrm{P} \text { is equal to } 1 : 81}$

Solution:

The expression for constant $\mathrm{K}_{\mathrm{p}}$ of equilibrium is given by:

$K_{p}=\frac{\text { Partial Pressure of } \mathrm{NO} \times \text { Partial Pressure of } \mathrm{Br}_{2}}{\text { Partial Pressure of } 2 \mathrm{NOBr}}$

$K_{p}=\frac{(P N O)^{2} \times P B r_{2}}{P N O B r}$

According to the chemical reaction given, the pressure of 2NO will be two times more than that of Br_2 which means partial pressure of 2NO will be 2 \times \frac{P}{9}

We are also given that P is the total pressure.

This means,

$2NOBr =P-\left(2 \times \frac{P}{9}\right)-\left(\frac{P}{9}\right)=\frac{9 P-2 P-P}{9}=\frac{6 P}{9}=\frac{2 P}{3}$

We can now calculate the constant of equilibrium $\mathrm{K}_{\mathrm{p}}$

$K_{p}=\frac{\left(2 \times \frac{P}{9}\right)^{2} \times \frac{P}{9}}{\left(2 \times \frac{P}{3}\right)^{2}}$

$K_{p}=\frac{\left(4 \times \frac{P^{2}}{81}\right) \times \frac{P}{9}}{\frac{4 \times P^{2}}{9}}$

$=\frac{\frac{4 \times P^{2}}{729}}{\frac{4 \times P^{2}}{9}}$

$=\frac{4 \times P^{3}}{729} \times \frac{9}{4 \times P^{2}}$

$K_{p}=\frac{P}{81}$

The ratio of $K_p/P$ is given below:

$\Rightarrow \frac{\frac{P}{81}}{P}=\frac{P}{81} \times \frac{1}{P}=\frac{1}{81}$

So, we get the ratio of $\mathrm{K}_{\mathrm{p}} : \mathrm{P}=1 : 81$