Question

The water-gas shift reaction describes the reaction of carbon monoxide and water vapor to form carbon dioxide and hydrogen (the mixture of carbon monoxide and hydrogen is known as water gas):
CO(g)+H2O(g)→CO2(g)+H2(g)

This reaction is an important industrial reaction that is used in the manufacture of ammonia, hydrocarbons, methanol, and hydrogen.

Use the following information to calculate ΔH∘ in kilojoules for the water-gas shift reaction:

C(s)+O2(g)→CO2(g) ΔH∘=−393.5kJ
2CO(g)→2C(s)+O2(g) ΔH∘=+221.0kJ
2H2(g)+O2(g)→2H2O(g) ΔH∘=−483.6kJ
Express your answer to one decimal place and include the appropriate units.

Answer 1

Answer:

ΔH∘ = - 41.2 KJ

Explanation:

We want to obtain the change in enthalpy for the reaction

CO(g) + H₂O(g) → CO₂(g) + H₂(g) (Main reaction)

And we're given the heat of formation of the reactants and products in the reaction

C(s) + O₂(g) → CO₂(g) ΔH∘=−393.5kJ (Reaction A)

2CO(g) → 2C(s) + O₂(g) ΔH∘=+221.0kJ (Reaction B)

2H₂(g) + O₂(g) → 2H₂O(g) ΔH∘=−483.6kJ (Reactions C)

To achieve this, we use the Born-Haber cycle.

The Born-Haber cycle entails writing the change in enthalpy of a reaction as a sum of change in enthalpies of a number of reactions that sum up to give the reaction whose enthalpy we needed from the start.

The main reaction is a sum of a sort of combination of Reactions A, B and C. We find this combination now.

From the reactions whose change in enthalpies are given,

C(s) + O₂(g) → CO₂(g) ΔH∘=−393.5kJ (Reaction A)

2CO(g) → 2C(s) + O₂(g) ΔH∘=+221.0kJ (Reaction B)

Dividing through by 2

CO(g) → C(s) + (1/2)O₂(g) ΔH∘=+110.5kJ (the enthalpy is divided by 2 too)

This reaction becomes (Reaction B)/2

2H₂(g) + O₂(g) → 2H₂O(g) ΔH∘=−483.6kJ (Reactions C)

Changing the direction of the reaction

2H₂O(g) → 2H₂(g) + O₂(g) ΔH∘=483.6kJ (the sign on the change in enthalpy changes)

Then, dividing by 2

2H₂O(g) → 2H₂(g) + O₂(g) ΔH∘=+483.6kJ

H₂O(g) → H₂(g) + (1/2)O₂(g) ΔH∘=241.8kJ (the change in enthalpy is divided by 2 too)

This reaction becomes (-Reaction C)/2

But, now, our main reaction can be written as a sum of these new Reactions,

Main Reaction = (Reaction A) + [(Reaction B)/2] + [(- Reaction C)/2]

C(s) + O₂(g) + CO(g) + H₂O(g) → CO₂(g) + C(s) + (1/2)O₂(g) + H₂(g) + (1/2)O₂(g)

Which gives the main reaction after eliminating the O2 that appears on both sides.

CO(g) + H₂O(g) → CO₂(g) + H₂(g)

Hence,

(ΔH∘ for the main reaction) = (ΔH∘ for reaction A) + [(ΔH∘/2) for reaction B) - [(ΔH∘/2) for reaction C)

ΔH∘ = - 393.5 + (221/2) - (-483.6/2) = - 41.2 KJ