**When dealing with equilibrium:**

• **K _{c} **→ equilibrium units are in molarity

•

**K _{p} and K_{c} are related to one another by the following equation below: **

$\overline{){{\mathbf{K}}}_{{\mathbf{p}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{c}}}{\left(\mathbf{RT}\right)}^{\mathbf{\u2206}\mathbf{n}}}$

**Reaction:**

2SO_{2}(g) + O_{2}(g) ⇌ 2SO_{3}(g)

reactants = 3 moles gas

products = 2 moles gas

Δn = 2 mol – 3 mol**Δ****n = -2**

$\overline{){{\mathbf{K}}}_{{\mathbf{p}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{c}}}{\left(\mathbf{RT}\right)}^{\mathbf{\u2206}\mathbf{n}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\left(\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{\times}{\mathbf{10}}^{\mathbf{8}}\right){\left[\left(\mathbf{0}\mathbf{.}\mathbf{08206}\right)\left(\mathbf{600}\right)\right]}^{\mathbf{-}\mathbf{2}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\left(\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{\times}{\mathbf{10}}^{\mathbf{8}}\right)\left(\mathbf{4}\mathbf{.}\mathbf{13}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\right)$

**K _{p} = 7.0 x 10^{4}**

An engineer examining the oxidation of SO _{2} in the manufacture of sulfuric acid determines that K_{c} = 1.7 x 108 at 600 K:

2SO_{2}(g) + O_{2}(g) ⇌ 2SO_{3}(g)

(a) At equilibrium, PSO_{3} = 300. atm and PO_{2} = 100. atm. Calculate PSO_{2}.

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Our tutors have indicated that to solve this problem you will need to apply the Equilibrium Expressions concept. If you need more Equilibrium Expressions practice, you can also practice Equilibrium Expressions practice problems.

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

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Our data indicates that this problem or a close variation was asked in Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition. You can also practice Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition practice problems.