Question

What is the effect of halving the pressure by doubling the volume on the following system at \[500{}^\circ C\]?
${{H}_{{{2}_{(g)}}}}+{{I}_{{{2}_{(g)}}}}2H{{I}_{(g)}}$
(A) Shift towards the reactant side
(B) Shift towards the product side
(C) Liquefaction of $HI$
(D) No effect

Answer 1

Hint: In order to solve this question we should be familiar with the Le-Chatelier’s Principles as to how they explain different effects of volume, pressure, temperature, etc. on the shift of equilibrium.

Complete Step by Step Solution:
Chemical equilibrium refers to the situation in a chemical reaction where both the reactants and products are present in concentrations that have no further tendency to change over time, preventing any discernible change in the system's properties. But, equilibrium can be disturbed by changing different factors like pressure, volume and temperature. There's a set of principles or rules called Le-Chatelier's principle which deals with the shift of equilibrium with change in different factors.

According to Le-chatelier’s principle, there is no effect on equilibrium if the volume has changed(no matter if the volume has decreased or the volume has increased). But on the contrary, there is an effect on changing the pressure. If we increase the pressure then the equilibrium shifts towards that side which has less number of moles of gaseous molecules.

Now, according to the given question, the equilibrium chemical reaction given to us is:
${{H}_{{{2}_{(g)}}}}+{{I}_{{{2}_{(g)}}}}2H{{I}_{(g)}}$ and given in the question is that we are decreasing the pressure and increasing the volume. Since volume has no effect on shift of equilibrium we’ll look for the effect of pressure. And we have seen as the pressure increases the equilibrium shift towards that side which has less number of moles.

Since, the pressure is decreasing it’ll move towards that side which has lesser number of moles but if we look at the given chemical reaction we can see that the number of gaseous moles is same on both the sides. Hence, there will be no effect of decreasing the pressure to half or doubling the volume.
Therefore, the correct option is D. No effect

Note: The question is completely dependent on the factors that changes the direction of equilibrium. Le-Chatelier’s principle are a set of rules which tell us the different effect of changing the different factors. The thing to remember in case of changing pressure is that the number of moles of only gaseous molecules should be taken into consideration and solid and liquid molecules have no role to play in this.