Question

If kp is the equilibrium constant for the reaction 2A(g) + B(g) ⇌ 3C(g), what is the equilibrium constant (in terms of kp) for the reaction 6A(g) + 3B(g) ⇌ 9C(g)?

Answer 1

The equilibrium constant for the tripled reaction 6A(g) + 3B(g) ⇌ 9C(g) is the original equilibrium constant Kp raised to the third power, resulting in Kp'^3.

Step-by-step explanation:

If we have the equilibrium constant Kp for the reaction 2A(g) + B(g) ⇌ 3C(g), we look to determine the equilibrium constant for the reaction when we triple the coefficients, resulting in the reaction 6A(g) + 3B(g) ⇌ 9C(g). The equilibrium constant for a reaction depends on the stoichiometry of the reaction. When the coefficients in the balanced equation are multiplied by a factor, the original equilibrium constant must be raised to the power of that factor.

Here, we have tripled the coefficients of the original reaction to get from 2A(g) + B(g) ⇌ 3C(g) to 6A(g) + 3B(g) ⇌ 9C(g), so we must raise Kp to the third power. Therefore, the new equilibrium constant Kp' for the tripled reaction will be Kp3.