Question

When each of the following equilibria is disturbed by increasing the pressure as a result of decreasing the volume, does the number of moles of reaction products increase, decrease, or remain the same?
1. CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
2. 2CO(g) ⇌ C(s) + CO2(g)
3. N2O4(g) ⇌ 2NO2(g)

Answer 1

Answer:

1. $CO(g) + H_2O(g)\rightleftharpoons CO_2(g) + H_2(g)$, remain the same.

2. $2CO(g)\rightleftharpoons C(s) + CO_2(g)$,increase.

3. $N_2O_4(g)\rightleftharpoons 2NO_2(g)$, decrease.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Increase the pressure

If the pressure of the container is increased, the volume will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place.

1. $CO(g) + H_2O(g)\rightleftharpoons CO_2(g) + H_2(g)$

On increasing the pressure ,the equilibrium will shift in direction where decrease in pressure is taking place,so reaction will shift in no direction because number of gaseous moles are same on both sides. Hence no increase in number of moles of products.

2. $2CO(g)\rightleftharpoons C(s) + CO_2(g)$

On increasing the pressure ,the equilibrium will shift in direction where decrease in pressure is taking place,so reaction will shift in forward direction because number of gaseous moles are less on product sides. Hence ,increase in number of moles of products.

3. $N_2O_4(g)\rightleftharpoons 2NO_2(g)$

On increasing the pressure ,the equilibrium will shift in direction where decrease in pressure is taking place,so reaction will shift in backward direction because number of gaseous moles are less on reactant side both sides. Hence, decrease in number of moles of products.