Question

Given K = 3.58 at 45°C for the reaction A(g) + B(g) equilibrium reaction arrow C(g) and K = 7.06 at 45°C for the reaction 2 A(g) + D(g) equilibrium reaction arrow C(g) what is the value of K at the same temperature for the following reaction? C(g) + D(g) equilibrium reaction arrow 2 B(g) What is the value of Kp at 45°C for the same reaction? Starting with 1.46 atm partial pressures of both C and D, what is the mole fraction of B once equilibrium is reached?

Answer 1

The important trick you need to solve the exercise is that the K of the sum of two reactions is the  the multiplication of each individual k. We also need to know that of you multiply an equilibrium constant by a factor this become

K´ = Kⁿ

A(g)  + B(g)    C(g) K1 = 3.58

2 A(g) + D(g)   C(g) K2 = 7.06

We need to multiply the first equation by 2 and reverse it-

2C(g) 2A(g)  + 2 B(g) K = 1/3.58² = 0.07802

2 A(g) + D(g)   C(g) K2 = 7.06

sum

C(g) + D(g)     2B(g) Kc = 0.07802 x 7.06 = 0.5508

2) Kp = Kc(RT) but as = 0 (because you have 2 moles of B - 1 mol of C - 1 mol of D = 0)

Kp= Kc = 0.5508

3)