Question

Following are three reaction given
$A_{\left(g\right)}+B_{\left(g\right)}\rightleftharpoons C_{\left(g\right)}+D_{\left(g\right)}$
$A_{\left(g\right)}\rightleftharpoons C_{\left(g\right)}+D_{\left(g\right)}$
$A_{\left(g\right)}+B_{\left(g\right)}\rightleftharpoons D_{\left(g\right)}$
Which of the following statements are correct?

• A. lncrease of pressure shifts the equilibrium II to the left.
• B. Decrease in volume shifts the equilibrium III to the right.
• C. Addition of He at constant volume shifts equilibrium II to the left.
• D. Addition of He at constant pressure shifts the equilibrium III to the left.

Answer 1

Answer: (A), (B), (D)

Addition of He at constant pressure shifts the equilibrium III to the left.
lncrease of pressure shifts the equilibrium II to the left.
Decrease in volume shifts the equilibrium III to the right.

When pressure is increased, the equilibrium will shift in a direction in which there is decrease in the number of moles. Due to this, the pressure will decrease and nullify the increase in the pressure. In reaction II, the number of moles of products is greater than the number of moles of reactant. Hence, increase of pressure shifts the equilibrium II to the left. In reaction III, the number of moles of reactants is greater than the number of moles of product. When volume is decreased pressure is increased. Hence, decrease in volume shifts the equilibrium III to the right. Addition of He at constant volume has no effect on the position of the equilibrium. When inert gas is added at constant presure, the equilibrium will shift in the direction in which there is increase in the number of moles of gases. When He is added at constant pressure, the equilibrium of reaction II will shift to left.