Chemistry 102: Equilibrium: Calculate the equilibrium constant

Question

Given the following reactions:

3CD  KC​=6.12×10−4

D3A+3B  KC​=7.14×10−2

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

A+BC

 

Answer 1

Hess' Law

Given:

eq.1: 3C <==> D ... K = 6.12x10^-4

eq.2: D <==> 3A + 3B ... K = 7.14x10^-2

Asked for:

A + B <==> C ... K = ?

Procedure:

Add the two together and combine/cancel like terms

3C + D <==> D + 3A + 3B

3C <==> 3A + 3B ... K = (6.12x10^-4)(7.14x10^-2) = 4.37x10^-5

Reverse and divide by 3 to get ...

A + B <==> C which is the desired reaction.

Since we reversed the reaction and divided by three, we have to treat the K as folllows:

K' = (1 / 4.37x10^-5)^-3

K' = 28.4