Question

Which of the following equilibria would shift right by increasing the volume of the reaction vessel based on Le Châtelier's principle?

(a) N 2 (g)+3H 2 (g) = 2NH 3 (g)

(b) H 2 (g)+l 2 (g)=2Hl(g)

(c) ( CO(g)+H 2 O(g) = CO 2 (g)+H 2 (g)

(d) CaO(s)+CO 2 (g) = CaCO 3 (s)

(e) N 2 O 4 (g)=2NO 2 (g)

Answer 1

Option e, is correct answer

Increased volume shifts the system to the side of the reaction that has more moles of gas,  whereas, decreased volume shifts the system to the side of the reaction that has fewer moles of gas.

Here, we have to increase the volume, then, for shifting the equilibria on right side, i.e., formation of product, would occur if initially, there are more number of moles on right side, product side.

in option e, there are 2 moles in product side, and 1 mole in reactant side, so here, increase of volume,will shift the equilibria on right side, i.e., formation of product.

In options, a and d, number of moles are decreasing on product side, so increase in volume will shift equilibria on reactant side.

In options, b and c , reaction has the same number of moles of gaseous reactants and products, so changing the volume for the reaction will not shift the system either way. Neither reactants nor products are favored.