In which of the following reactions does a decrease in the volume of the reaction vessel at constant

temperature favor formation of the products?

A) 2H2 (g) + O2 (g) - -> 2H2O (g)

B) NO2 (g) + CO (g) - -> NO (g) + CO2 (g)

C) H2 (g) + I2 (g) - -> 2HI (g)

D) 2O3 (g) - -> 3O2 (g)

E) MgCO3 (s) - -> MgO (s) + CO2 (g)

The correct option is: A) 2H₂ (g) + O₂ (g) → 2H₂O (g)

Explanation:

According to the Le Chatelier's principle, change in the volume of the reaction system causes equilibrium to shift in the direction that reduces the effect of the volume change.

When the volume decreases then the pressure of the reaction vessel increases, then the equilibrium shifts towards the reaction side that produces less number of moles of gas.

A) 2H₂ (g) + O₂ (g) → 2H₂O (g)

The number of moles of reactant is 3 and number of moles of product is 2.

Therefore, when volume decreases, the equilibrium shifts towards the product side, thereby favoring the formation of products.

B) NO₂ (g) + CO (g) → NO (g) + CO₂ (g)

The number of moles of reactant and product both is 2.

Therefore, when the volume decreases, the equilibrium does not shift in any direction.

C) H₂ (g) + I₂ (g) → 2HI (g)

The number of moles of reactant and product both is 2.

Therefore, when the volume decreases, the equilibrium does not shift in any direction.

D) 2O₃ (g) → 3O₂ (g)

The number of moles of reactant is 2 and number of moles of product is 3.

Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.

E) MgCO₃ (s) → MgO (s) + CO₂ (g)

The number of moles of reactant is 1 and number of moles of product is 2.

Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.