Question

Consider the equilibrium system described by the chemical reaction below. If the partial pressures at equilibrium of NO, Cl₂, and NOCl are 0.095 atm, 0.171 atm, and 0.28 atm, respectively, in a reaction vessel of 7.00 L at 500 K, what is the value of Kp for this reaction? 2 NO(g) + Cl₂(g) ⇌ 2 NOCl(g)

Answer 1

To find the equilibrium constant Kp for the reaction 2 NO(g) + Cl₂(g) ⇌ 2 NOCl(g), you use the given equilibrium partial pressures to calculate Kp, which in this case is approximately 50.846 atm.

Step-by-step explanation:

To calculate the equilibrium constant (Kp) for the reaction given the partial pressures at equilibrium, we use the expression for Kp which is based on the partial pressures of the gaseous components. For the reaction 2 NO(g) + Cl₂(g) ⇌ 2 NOCl(g), the equilibrium expression is:

Kp = (P_NOCl)^2 / (P_NO)^2 * P_Cl2

Plugging in the values given, we get:

Kp = (0.28 atm)^2 / ((0.095 atm)^2 * 0.171 atm)

To solve for Kp, we do the math:

Kp = (0.0784 atm^2) / (0.009025 atm^2 * 0.171 atm)

Kp = (0.0784 / (0.00154225 atm^3))

Kp = 50.846 atm

Therefore, the value of Kp for this equilibrium reaction at 500 K is approximately 50.846.