Question

Two reactions and their equilibrium constants are given.


A+2B−⇀2C K1=2.93


2C−⇀D K2=0.258



Calculate the value of the equilibrium constant for the reaction


D−>A+2B

.

Answer 1

Adding the two equations gives:

A + 2B = 2C K₁=2.93

+

2C = D K₂=0.258

Result:

A + 2B + 2C = 2C + D

K₁ + K₂

Remove 2С:

A + 2B = D

K₁ + K₂ = 2.93 + 0.258 = 3.188

Reversed reaction:

D = A + 2B

K₃ = 1 / 3.188 = 0.314