Question

Question: Consider the reaction

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

At 175⁰C and a pressure of 128 torr, an equilibrium mixture of O₂ and O₃ has a density of 0.168 g/L. Calculate K_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

Given data

$\begin{array}{rcl}Density& =& \rho \\ & =& \frac{P\times \left(molarmass\right)}{RT}\\ & =& \frac{P_{O_2}\times \left(molarmassof{O}_2\right)+P_{O_3}\times \left(molarmassof{O}_3\right)}{RT}\\ 0.168g/L& =& \frac{P_{O_2}\times \left(32.00g/mol\right)+P_{O_3}\times \left(48.00g/mol\right)}{0.08206{\displaystyle \frac{Latm}{Kmol}}\times 448K}\\ 6.18& =& P_{O_2}\times \left(32.00\right)+P_{O_3}\times \left(48.00\right)\end{array}$

Now from the given data we can write

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

Determine KP

Solving equation $\begin{array}{rcl}32P_{O_2}+48P_{O_3}& =& 6.18\end{array}$ and $\begin{array}{rcl}{P}_{O_2}+P_{O_3}& =& 0.168\end{array}$ we get

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

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

$\begin{array}{rcl}{K}_P& =& \frac{{P_{O_3}}^2}{{P_{O_2}}^3}\\ & =& \frac{{\left(0.05\right)}^2}{{\left(0.118\right)}^3}\\ & =& 1.5at{m}^{-1}\end{array}$