Question

Suppose a particular mechanism has the following equilibria:

Ni(CO)_4(g) ⇌ Ni_(s) + 4 CO_(g)               K₁ = 8.77             ΔH₁ = 190 kJ

MnO_(s) + CO_(g) ⇌ Mn_(s) + CO_2(g)        K₂ = 490              ΔH₂ = 349 kJ

Mn_(s) + H₂O_(g) ⇌ MnO_(s) + H_2(g)         K₃ = 0.0149          ΔH₃ = -346 kJ

Determine the values of the equilibrium constant and enthalpy change for the net equation:

Ni(CO)_4(g) + 4 H₂O_(g) ⇌ Ni_(s) + 4 CO_2(g) + 4H_2(g)        K_c = ?     ΔH = ?

(You’ll want to review the rules for manipulating the stoichiometry of an equilibrium reaction and its equilibrium constant in section 14.3 of the textbook. It is different than using Hess’s Law to find the net enthalpy change.)

3 species do not appear in the net question. Which are they? Classify each as an intermediate or a catalyst. How do you know the difference?

Answer 1

Ni(CO)4(g) ⇌ Ni(s) + 4 CO(g)               K1 = 8.77             ΔH1 = 190 kJ

MnO(s) + CO(g) ⇌ Mn(s) + CO2(g)        K2 = 490              ΔH2 = 349 kJ

Mn(s) + H2O(g) ⇌ MnO(s) + H2(g)         K3 = 0.0149          ΔH3 = -346 kJ

To calculate Kc for equilibrium reaction

Ni(CO)4(g) + 4 H2O(g) ⇌ Ni(s) + 4 CO2(g) + 4H2(g)

we need to take

eq + 4 x eq 2 + 4 x eq 3

When we multiply an equation by a factor x , the K vlue becomes K^x and delta H is multiplied by x .and addition of equations K values are multiplied, delta H values are added.

Thus the K for new equilibrium is

K new = K_1.(K₂)⁴(K3)⁴

   ^=8.77x(490)4(0.0149)⁴

   = 24918

Delta H new = 190 + 4(349) + 4( -346 )

= 202kJ

The species that do not appear in the net equation are Mn, MnO, and CO.

The catalyst is MnO as it is reacted and regenerated in the other step.

The intermediates are Mn and CO