Question

Consider this reaction at equilibrium at a total pressure P1:

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

Suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished. The new equilibrium total pressure will be

A) twice P1

B) three times P1

C) 3.5 P1

D) less than twice P1

E) unchanged

The answer is D, but would someone solve this question step-by-step?

Answer 1

Volume is reduce to one half

Final volume initial volume.

Using relationship

We can say that

Note: Here we have substituted

Hence

Here and are initial pressure and initial volume respectively. and are final pressure and final volume respectively.

Hence, final total pressure will be twice the initial total pressure.

But now we will apply Le chatelier's principle. We have increased total pressure. The system will try to nullify the effect of increase in pressure. This will happen when more and more SO3 (product) is formed. This decreases the number of moles of gaseous species and decreases pressure. Hence, the new equilibrium total pressure will be less than twice .

Note: Total number of moles of gaseous products are less than total number of moles of gaseous reactants. When pressure is increased, the equilibrium will shift in forward direction.