# Problem: An engineer examining the oxidation of SO 2 in the manufacture of sulfuric acid determines that Kc = 1.7 x 108 at 600 K: 2SO2(g) + O2(g) ⇌ 2SO3(g) (a) At equilibrium, PSO3 = 300. atm and PO2 = 100. atm. Calculate PSO2.

🤓 Based on our data, we think this question is relevant for Professor Randles' class at UCF.

###### FREE Expert Solution

When dealing with equilibrium:

Kc → equilibrium units are in molarity
Kp → equilibrium units in terms of pressure

Kp and Kc are related to one another by the following equation below:

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

Reaction:

2SO2(g) + O2(g) ⇌ 2SO3(g)

reactants = 3 moles gas

products = 2 moles gas

Δn = 2 mol – 3 mol
Δn = -2

$\overline{){{\mathbf{K}}}_{{\mathbf{p}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{c}}}{\left(\mathbf{RT}\right)}^{\mathbf{∆}\mathbf{n}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\left(\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{×}{\mathbf{10}}^{\mathbf{8}}\right){\left[\left(\mathbf{0}\mathbf{.}\mathbf{08206}\right)\left(\mathbf{600}\right)\right]}^{\mathbf{-}\mathbf{2}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\left(\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{×}{\mathbf{10}}^{\mathbf{8}}\right)\left(\mathbf{4}\mathbf{.}\mathbf{13}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\right)$

Kp = 7.0 x 104

###### Problem Details

An engineer examining the oxidation of SO 2 in the manufacture of sulfuric acid determines that Kc = 1.7 x 108 at 600 K:

2SO2(g) + O2(g) ⇌ 2SO3(g)

(a) At equilibrium, PSO3 = 300. atm and PO2 = 100. atm. Calculate PSO2