An engineer examining the oxidation of SO₂ in the manufacture

of sulfuric acid determines that${\mathbf{K}}_{\mathbf{C}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}$at 600. K:

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

(a) At equilibrium,${\mathbf{P}}_{{\mathbf{SO}}_{\mathbf{3}}}\mathbf{=}\mathbf{300}\mathbf{.}\mathbf{atm}$and${\mathbf{P}}_{{\mathbf{O}}_2}\mathbf{=}1\mathbf{00}\mathbf{.}\mathbf{atm}$.Calculate${\mathbf{P}}_{{\mathbf{SO}}_{\mathbf{2}}}$

(b) The engineer places a mixture of 0.0040 mol of SO₂(g) and 0.0028 mol of O₂(g) in a 1.0-L container and raises the temperature to 1000 K. At equilibrium, 0.0020 mol of SO₃(g) is present. Calculate Kc and for this reaction at 1000. K.

Given balance chemical equation is shown below:

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

Given data:${\text{K}}_{\text{C}}{\text{= 1.7×10}}^{\text{8}}$

${\text{P}}_{{\text{SO}}_{\text{3}}}\text{= 300.atm}$

${\text{P}}_{{\text{O}}_{\text{2}}}\text{= 100.atm}$

To find the partial pressure of SO₂, equation in terms of Kp is defined as,

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

We need to find the value of kp using Kc given above:

${\mathbf{K}}_{\mathbf{P}}\mathbf{=}{\mathbf{K}}_{\mathbf{c}}{\left(\mathbf{RT}\right)}^{\mathbf{∆}\mathbf{n}}$ (2)

Were,$∆n=2-3=-1$

${\text{K}}_{\text{P}}{\text{= 1.7×10}}^{\text{8}}\text{×}{\left(\text{0.082×1600}\right)}^{\text{- 1}}$

${\text{= 3.451×10}}^6\text{atm}$

Substituting the value of Kp in equation 1 we get,

${\text{3.451×10}}^{\text{6}}\text{atm =}\frac{{\left[\text{300.atm}\right]}^{\text{2}}}{{\left[{\text{P}}_{{\text{SO}}_{\text{2}}}\right]}^{\text{2}}\left[\text{100.atm}\right]}$

${\left[{\text{P}}_{{\text{SO}}_{\text{2}}}\right]}^{\text{2}}\text{=}\frac{{\left[\text{300.atm}\right]}^{\text{2}}}{{\text{3.451×10}}^{\text{6}}\text{atm×}\left[\text{100.atm}\right]}$

$\text{= 2.6079 atm}$

$$\begin{gathered} \left[{\text{P}}_{{\text{SO}}_{\text{2}}}\right]\text{=}\sqrt{\text{2.6079}} \\ \text{= 1.6149 atm} \end{gathered}$$

Equilibrium constant, ${\text{K}}_{\text{C}}\text{=}\frac{{\left[{\text{SO}}_{\text{3}}\right]}^{\text{2}}}{{\left[{\text{SO}}_{\text{2}}\right]}^{\text{2}}\left[{\text{O}}_{\text{2}}\right]}$

$\text{=}\frac{{\left(\text{0.0020}\right)}^{\text{2}}}{{\left(\text{0.0040}\right)}^2\left(\text{0.0028}\right)}$

${\text{= 7.0×10}}^{\text{4}}$

Thus, Kc is calculated, and it is${\text{7.0×10}}^{\text{4}}$

Partial pressure ossf the reactant SO₂ can be determined using the ideal gas law given below:

$$\begin{gathered} \text{PV = nRT} \\ \text{ie, P =}\frac{\text{n}}{\text{v}}\text{RT} \end{gathered}$$

${\text{P}}_{{\text{SO}}_{\text{2}}}\text{= 0.0040}\frac{\text{mol}}{\text{L}}\text{×0.0821}\frac{\text{L. atm}}{\text{mol. k}}\text{×1000K}$

$\text{= 0.3284 atm}$

Hence partial pressure of SO₂ is 0.3284atm