Question

Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2NH3(g)+ 3N20(g)- 4N2(g)+ 3H20(g) ANSWER kJ

Answer 1

1.Ans :- ΔH⁰ = - 878.4 KJ

Explanation :-

Given reaction is :

2 NH₃ (g) + 3 N₂O (g) ------------> 4 N₂ (g) + 3 H₂O (g) , ΔH⁰ = ?

As,

ΔH⁰ = [Sum of enthaply of formation (Δ_fH⁰) of products ] - [Sum of enthalpy of formation (Δ_fH⁰​​​​​​​) of reactants ]

ΔH⁰ = [4xΔ_fH⁰ of N₂(g) + 3xΔ_fH⁰ of H₂O​​​​​​​(g)] - [2xΔ_fH⁰ of NH₃(g) + 3xΔ_fH⁰ of N₂​​​​​​​O​​​​​​​(g)]

ΔH⁰ = [4x0.0 KJ/mol + 3x(–241.8 KJ/mol)] - [2x(–45.9 KJ/mol) + 3x(81.6 KJ/mol)]

ΔH⁰ = [- 725.4 KJ ] - [ - 91.8 KJ + 244.8 KJ]

ΔH⁰ = - 725.4 KJ - 153.0 KJ

ΔH⁰ = - 878.4 KJ

Hence, ΔH0 = - 878.4 KJ

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2.Ans :- ΔH⁰ = - 91.8 KJ

Explanation :-

Given reaction is :

N₂ (g) + 3 H₂ (g) ------------> 2 NH₃ (g) , ΔH⁰ = ?

As,

ΔH⁰ = [Sum of enthaply of formation (Δ_fH⁰​​​​​​​) of products ] - [Sum of enthalpy of formation (Δ_fH⁰​​​​​​​) of reactants ]

ΔH⁰ = [2xΔ_fH⁰ of NH₃(g)] - [1xΔ_fH⁰ of N₂ (g) + 3xΔ_fH⁰ of H₂​​​​​​​(g)]

ΔH⁰ = [2x(–45.9 KJ/mol)] - [1x0.0KJ/mol + 3x0.0 KJ/mol]

ΔH⁰ = [- 91.8 KJ ] - [0.0 KJ]

ΔH⁰ = - 91.8 KJ

Hence, ΔH0 = - 91.8 KJ

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