Calculate the value of the equilibrium constant for the following reaction.

Question

Given:2C <==> D ... Kc = 6.52x10^-4

Given:D <===> 3A + 3B ... Kc = 4.32x10^-2

A + B <==> C

Answer 1

Here's some rules you should remember with respect to the effect of manipulating reactions on the equilibrium constants:

1. Reversing the direction of the reaction causes the new equilibrium constant (K') to be the inverse of the original (K). K' = 1/K
2. Multiplying the reaction by a coefficient n (for dividing by x, think of multiplying by 1/x) causes the new equilibrium constant to be equal to the original equilibrium constant (K) raised to the nth power. K' = Kⁿ
3. The equilibrium constant for the sum of two reactions (K3) is equal to the product of the equilibrium constants for each of the individual reactions (K1, K2). K3 = K1*K2.

Following these rules, you should be able to get the answer pretty straightforwardly. For a step by step solution, please feel free to book some time with me!

Answer 2

We will use Hess's Law for this problem. We want to manipulate the given equations such that when they are added together, we obtain the target equation. Assuming the 1st given reaction is actually 3C <==> D and not 2C <==> D, we proceed as follows:

TARGET EQUATION: A + B <==> C

Given:

eq. 1. 3C <==> D ... Kc = 6.52x10^-4

eq. 2. D <===> 3A + 3B ... Kc = 4.32x10^-2

Procedure:

reverse eq. 1 and divide by 3. This gets C on the right side of the equation:

1/3 D <==> C and K' = 1 / (Kc)^1/3 = 1 / (6.52x10^-4)^1/3 = 1 / 0.0867 = 11.53

reverse eq. 2 and divide by 3. This gets A and B on the left side of the equation:

A + B <==> 1/3 D and K' = 1 / (Kc)^1/3 = 1 / (4.32x10^-2)^1/3 = 2.85

add these 2 equations together to get:

1/3 D + A + B <==> C + 1/3 D which reduces to the target equation of A + B <==> C.

When adding 2 equations, we multiply the 2 corresponding K' values to get the new K:

Kc _{(A+B <==>C)} = 11.53 x 2.85 = 32.9