Question

The gas stream from a sulfur burner is composed of 15-mol% SO2, 20-mol% O2, and 65-mol% N2. This gas stream at 1 bar and 480?C enters a catalytic converter, where the SO2 is further oxidized to SO3. Find the equilibrium conversion of SO2. The equilibrium constant for the reaction

Answer 1

Answer: the equilibrium conversion is 0.15 or 15mole %

K=1.33

Explanation:

Given the mole percent of the gases to be as follows

SO2(g) =15% = 0.15mol

O2(g) =20% = 0.20mol

N2(g) = 65% = 0.65mol

The equilibrium conversion reaction is given by

2SO2(g) + O2(g) <------> 2SO3(g)

2 1. 2

According to the reaction 2 moles of SO2 is converted to get 2 mole of SO3

Therefore 0.15 mole will produce 0.15 mole of SO3

So at equilibrium:

2SO2(g) + O2(g) ------> 2SO3(g)

0.15. 0.075. 0.15

Now the equilibrium constant K is given by:

K = [SO3]²/[SO2]²[O2]

Since the container is constant for all the reactants we can neglect the container capacity and use the mole as percent concentration.

K = [0.15]²/[0.15]²[0.075]

K = 1/0.075

K= 1.33