Question

A sample of iron (II) sulfate was heated in an evacuated container to920 K, where the following reactions occurred:

$\mathbf{2}{\mathbf{FeSO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\rightleftharpoons }{\mathbf{FeSO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{SO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{SO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

${\mathbf{SO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathbf{SO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

After reaching equilibrium, the total pressure was 0.836 atm, and the partial pressure of oxygen was 0.0275 atm. Calculate Kp for each of the above reactions.

Short answer

The value of KP for chemical reaction

$2{\mathrm{FeSO}}_4\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{FeSO}}_3\left(\mathrm{s}\right)+{\mathrm{SO}}_3\left(\mathrm{g}\right)+{\mathrm{SO}}_2\left(\mathrm{g}\right)$is ${\mathrm{K}}_{\mathrm{P}}={(0.161)}^2$

And for reaction ${\mathrm{SO}}_3\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{SO}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)$is K_P = (0.218)^1/2

Step-by-step solution

As we have seen in the above reaction, the first reaction produces equal amounts of SO3 and SO2. 

So, by using the second reaction, we first calculate the partial pressures at the equilibrium of SO₃, SO₂, and O₂as follows:

${\mathbf{SO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathbf{SO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$.

Initial pressure P₀ P₀ 0

Change in pressure -x+x + x/2

At equilibrium (P_{0 -}x) P₀+x+x/2

${\mathrm{P}}_{\mathrm{total}}={\mathrm{P}}_0-\mathrm{x}+{\mathrm{P}}_0+\mathrm{x}+\frac{\mathrm{x}}{2}=0.836\mathrm{atm}$

${\mathrm{P}}_{{\mathrm{O}}_2}=\frac{\mathrm{x}}{2}=0.0275\mathrm{atm}$

x = 0.0550 atm

Now, by putting the value of ‘x’ we can get partial pressure of all gases in the reaction:

2P₀ + x/2 = 0.836 atm;

2P₀= 0.836 – 0.0275 = 0.809 atm

P₀= 0.405 atm

${\mathrm{P}}_{{\mathrm{SO}}_3}={\mathrm{P}}_0-\mathrm{x}=0.405-0.0550=0.350\mathrm{atm}$

${\mathrm{P}}_{{\mathrm{SO}}_2}={\mathrm{P}}_0+\mathrm{x}=0.405+0.0550=0.460\mathrm{atm}$

Now, put the values of partial pressure, and we can get the Kp value as follows:

For the first reaction: $\mathbf{}\mathbf{}\mathbf{2}{\mathbf{FeSO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\rightleftharpoons }{\mathbf{FeSO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{SO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{SO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

$\begin{array}{l}{\mathbf{K}}_{\mathbf{P}}\mathbf{=}{\mathbf{P}}_{{\mathbf{SO}}_{\mathbf{2}}}\mathbf{\times }{\mathbf{P}}_{{\mathbf{SO}}_{\mathbf{3}}}\\ {\mathbf{K}}_{\mathbf{P}}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{460}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{350}\mathbf{)}\\ {\mathbf{K}}_{\mathbf{P}}\mathbf{=}{\left(0.161\right)}^{\mathbf{2}}\end{array}$

For the second reaction:${\mathbf{SO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathbf{SO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

${\mathbf{K}}_{\mathbf{P}}\mathbf{=}\frac{\mathbf{\left(}{\mathbf{P}}_{{\mathbf{SO}}_{\mathbf{2}}}\mathbf{\right)}\mathbf{\times }{\mathbf{\left(}{\mathbf{P}}_{{\mathbf{O}}_{\mathbf{2}}}\mathbf{\right)}}^{\frac{\mathbf{1}}{\mathbf{2}}}}{\mathbf{\left(}{\mathbf{P}}_{{\mathbf{SO}}_{\mathbf{3}}}\mathbf{\right)}}$

${\mathrm{K}}_{\mathrm{P}}=\frac{(0.460)\times {(0.0275)}^{\frac{1}{2}}}{(0.350)}={(0.218)}^{\frac{1}{2}}\mathrm{atm}$

Hence the final answer is Kp of the first reaction:${\mathrm{K}}_{\mathrm{P}}={(0.161)}^2$

And for the second rection: ${\mathrm{K}}_{\mathrm{P}}={(0.218)}^{1/2}$