Question

A reaction vessel at 1215k contains a mixture of SO2( P = 2. 98 bar) and O2(P = 1. 18 bar). When a catalyst is added, this reaction takes place: 2SO2(g)+O2(g)⇌2SO3(g) At equilibrium the total pressure is 3. 80 bar.


Find the value of k?

Answer 1

This problem is providing the equilibrium reaction whereby SO3 is produced and asks for the value of the equilibrium constant. After the calculations, the answer turns out to be 0.124.

Chemical equilibrium.

In chemistry, when a chemical reaction attempts to go to completion, it might experience a limitation known as equilibrium; a condition where it is not able to proceed any further.

In such a way, for the given reaction:

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

We can write the equilibrium expression as:

$K=\frac{p_{SO_3}^2}{p_{SO_2}^2p_{O_2}}$

Hence, given the total pressure at equilibrium, we can add the pressures of all the species at equilibrium including the reaction extent, x:

$P_{eq}=p_{SO_3}+p_{SO_2}+p_{O_2}\\\\3.80=(2x)+(2.98-2x)+(1.18-x)$

Thus, we solve for x to obtain:

$3.80-2.98-1.18=2x-2x-x\\\\-0.36=-x\\\\x=0.36bar$

Hence, we plug it in into the equilibrium expression to obtain the equilibrium constant as follows:

$K=\frac{(2x)^2}{(2.98-2x)^2(1.18-x)}\\\\K=\frac{(2*0.36)^2}{(2.98-2*0.36)^2(1.18-0.36)}\\\\K=0.124$

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