Question

Consider the following reaction 2NOCl(g) → 2NO(g) + Cl₂(g)

Intitially pure NOCl(g) is placed in a vessel at 3.00 atm. At equilibrium, 0.416% of the NOCl is decomposed. Determine the value for Kp

Answer 1

The equilibrium constant (Kp) for the reaction 2NOCl(g) → 2NO(g) + Cl₂(g) is calculated to be 4.364 x 10^-6 using the partial pressures of the reactants and products at equilibrium.

Step-by-step explanation:

To calculate the equilibrium constant (Kp) for the reaction 2NOCl(g) → 2NO(g) + Cl₂(g), given that initially pure NOCl(g) is placed in a vessel at 3.00 atm, and at equilibrium, 0.416% of the NOCl is decomposed, we need to use the partial pressures of the products and reactants at equilibrium in the expression for Kp.

First, let's determine the equilibrium partial pressures:

• Initial pressure of NOCl = 3.00 atm
• 0.416% of NOCl decomposed = 0.00416 x 3.00 atm = 0.01248 atm
• At equilibrium, pressure of NOCl = initial pressure - decomposed = 3.00 atm - 0.01248 atm = 2.98752 atm
• Since the reaction decomposes 2 moles of NOCl for every 1 mole of Cl₂ produced, the pressure of Cl₂ = 0.5 x decomposed NOCl pressure = 0.5 x 0.01248 atm = 0.00624 atm
• Pressure of NO = pressure of Cl₂ (since they are produced in a 1:1 ratio) = 0.00624 atm

The expression for Kp is:

Kp = (PNO)² * PCl₂ / (PNOCl)²

Plugging the values into the formula:

Kp = (0.00624 atm)² * (0.00624 atm) / (2.98752 atm)²

Kp = 0.0000389504 atm³ / 8.92560430 atm²

Kp = 4.364 x 10-6