Question

What is the delta H of the following reaction of methane and ammonia in kJ/mole?

CH₄ (g) + NH₃ (g) -----> HCN (g) + 3H₂ (g) Delta H rxn = ? kJ / mole

Note, this is at a temperature different than the standard tables so you can't use those to solve this. Use the following reaction enthalpy data.

N₂ (g) + 3 H₂ (g) ---> 2 NH₃(g). Delta H = -101.8 kJ/mole

C (s) + 2 H₂ (g) ---> CH₄ (g) Delta H = -77.9 kJ/mole

H₂ (g) + 2 C (s) + N₂ (g) ----> 2 HCN (g) Delta H = +275.8 kJ/mol

Answer 1

For the given reaction:

CH4 (g) + NH3 (g) = HCN (g) + 3H2 (g),  ΔHrxn = ? kJ / mol

The given reactions are:

N2 (g) + 3 H2 (g) = 2 NH3 (g). ΔH= -101.8 kJ/mol ------(1)

C (s) + 2 H2 (g) = CH4 (g) ΔH= -77.9 kJ/mol -------------(2)

H2 (g) + 2 C (s) + N2 (g) = 2 HCN (g) ΔH = +275.8 kJ/mol-----(3)

Inverting equation 1 and 2, we get above three equation as:

2 NH3 (g) = N2 (g) + 3 H2 (g) , ΔH= +101.8 kJ/mol ----(1)

CH4 (g) = C (s) + 2 H2 (g), ΔH= +77.9 kJ/mol -----------(2)

H2 (g) + 2 C (s) + N2 (g) = 2 HCN (g) ΔH = +275.8 kJ/mol-----(3)

On dividing equations 1 and 3 by 2 , we get,

NH3 (g) = 1/2 N2 (g) + 3/2 H2 (g) , ΔH= +50.9 kJ/mol ----(1)

CH4 (g) = C (s) + 2 H2 (g), ΔH= +77.9 kJ/mol -----------(2)

1/2H2 (g) + C (s) + 1/2 N2 (g) = HCN (g) ΔH = +137.9 kJ/mol-----(3)

Now on adding above three equations and cancelling out the similar terms,we get;

CH4 (g) + NH3 (g) = HCN (g) + 3 H2 (g),  

ΔHrxn = +50.9 kJ/mol + +77.9 kJ/mol  +137.9 kJ/mol = +266.7 KJ