Question

At equilibrium in a 14.6 L vessel, there are 7.0×10−2 moles of SO2, 9.0×10−2 moles of O2, and 8.5×10−2 of SO3. What is the equilibrium constant (Kc) for the reaction: 2SO2(g) + O2(g) ⇌ 2SO3(g)

A. 16.4

B. 13.5

C. 239

D. 4.2×10−3

I attempted to do this with the moles as concentrations and I got answer A. It was wrong so I just want to know what other step I am missing and why do I have to convert? Is it because we are given volume of the vessel? Any help will be appreciated! Thank you in advance.

Answer 1

Given chemical transformation,

2SO₂(g) + O₂(g) <----------> 2SO₃(g).

For this chemical transformation we are suppllied with the number of moles of each species i.e. gases preesent in the vessel. Gases occupies all available space and hence these number of moles of gase are spread over toatal available volume i.e. volume of vessel.

For this transformation the equilibrium constant will be given as,

Kc = [SO₃]² / [SO₂]²[O₂]....................(1)

[SO₃], [SO₂], [O₂] represent concentrations of respective species in mol / L unit (also called active mass).

We are given with moles of gases in 14.6 L of vessel. i.e. these are moles of gases per 14.6 L. We must divide these numbers of moles by 14.6 to have the concentrations in the unit of mole/L.

[SO₃] = [(8.5 x 10^-2) / 14.6] = 5.82 x 10^-3 mole/L.

[SO₂] = [(7.0 x 10^-2) / 14.6] = 4.79 x 10^-3 mole/L.

[O₂] = [(9.0 x 10^-2) / 14.6] = 6.16 x 10^-3 mole/L.

Let us put these concentrations in eequation (1)

Kc = [5.82 x 10^-3]² / {[4.79 x 10^-3]^{2 x} [6.16 x 10^-3]}

Kc = (3.38 x 10^-5) / (2.29 x 10^-5) x (6.16 x 10^-3)

Kc = (3.38) / (2.29 x 6.16 x 10^-3).......................( 10^-5 factor cancelled mutually)

Kc = 239.6

Kc 239

Hence the equilibrium constant (Kc) for the given tansformation under given conditions is 239.

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