Question

Consider the reaction

$\mathbf{3}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{2}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

At 175⁰C and a pressure of 128 torr an equilibrium mixture of O₂ and O₃ has a density of 0.168 g/L. CalculateK_p for the above reaction at 175⁰C.

Short answer

K_p for the above reaction at 175⁰C is 1.5 atm^-1.

Step-by-step solution

Step 1:State the given data 

$\begin{array}{c}0.168g/L=\frac{P_{O_2}\times (32.00g/mol)+P_{O_3}\times (48.00g/mol)}{0.08206\frac{Latm}{Kmol}\times 448K}\\ 6.18=P_{O_2}\times \left(32.00\right)+P_{O_3}\times \left(48.00\right)\end{array}$

$\begin{array}{c}Density=\rho \\ =\frac{P\times \left(molarmass\right)}{RT}\\ =\frac{P_{O_2}\times \left(molarmassof{O}_2\right)+P_{O_3}\times \left(molarmassof{O}_3\right)}{RT}\end{array}$

Now from the given data we can write

$\begin{array}{c}{P}_{O_2}+P_{O_3}=128torr\\ =\frac{128torr\times 1atm}{760torr}\\ =0.168atm\end{array}$

Determine KP

Solving equation $6.18=32P_{O_2}+48P_{O_3}$ and ,$P_{O_2}+P_{O_3}=0.168$we get

$\begin{array}{l}{P}_{O_3}=0.05atm\\ P_{O_2}=0.118atm\end{array}$

Now, for the reaction given, the value of equilibrium constant will be

$\begin{array}{c}{K}_P=\frac{P_{O_3}^2}{P_{O_2}^3}\\ =\frac{{\left(0.50atm\right)}^2}{{\left(0.118atm\right)}^3}\\ =1.5at{m}^{-1}\end{array}$