Question

One way to represent this equilibrium is: 2 CO2(g) + 4 H2O(g)2 CH3OH(l) + 3 O2(g) We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above. 1) CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(g) K1 = 2) 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g) K2 = 3) CO2(g) + 2 H2O(g) CH3OH(l) + 3/2 O2(g)

Answer 1

Answer:

1) 1/√K

2) 1/K

3) √K

Explanation:

As general rules of operation of equilibrium constants:

For the reaction:

A + B ⇄ C + D; Equilibrium constant = K

C + D ⇄ A + B; K' = 1/K

2A + 2B ⇄ 2C + 2D; K'' = K²

Thus, as equilibrium constant of:

2CO2 + 4H2O ⇄ 2CH3OH + 3O2

Is K:

1) CH3OH + 3/2 O2 ⇄ CO2 + 2H2O

K' = 1 / K^(1/2) = 1/√K

2) 2CH3OH + 3O2 ⇄ 2CO2 + 4H2O

K' = 1/K

3) CO2 + 2H2O ⇄ CH3OH + 3/2 O2

K' = K^(1/2) = √K