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The standard enthalpy of formation of NH3 is -46.0 kJ mol-1. If the enthalpy of formation of H2 from its atoms is -436 kJ mol-1 and that of N2 is -712 kJ mol-1, the average bond enthalpy of N-H A ·n NH3 is ?

The standard enthalpy of formation of NH3 is -46.0 kJ mol-1. If the enthalpy of formation of H2 from its atoms is -436 kJ mol-1 and that of N2 is -712 kJ mol-1, the average bond enthalpy of N-H A ·n NH3 is ?

Grade:12

4 Answers

Ramreddy IIT Patna
askIITians Faculty 49 Points
10 years ago
deltaH= sum of enthalpy of formation of products- sum of enthalpy of formation of reactants
= (-2*46)- (-712 -3*436)
= 1928 kJ/mol
deltaH= sum of bond energies of reactants- sum of bond energies of products
1928 = (3*436+712) – (3*x)
we can calculate the x (Avg bond enegy of N-H) from above
Navish
12 Points
6 years ago
Given :
2(g) 2(g)
1 N + 3 H
2 2
°
3(g) g NH ;ΔH = –46
1 × 712
2
3 × 436
2
N + 3H (g) (g)
Average bond enthalpy of N–H bond = + 352
kJ mol.
Ankit
29 Points
6 years ago
Aree ya to answer ate nahi aur ate hai to ache se nahi ate faltu ask iitians 😢😢😢😢😢😢😭😭😭😛😛😛😠😠😠😠sale aaj tak bohot ans pooche par 2 ko chhodke aur kisi ka ans sahi se nahi diya. Adhe se jyada baar to ans hi nahi Mila 
Rishi Sharma
askIITians Faculty 646 Points
4 years ago
Dear Student,
Please find below the solution to your problem.

ΔH = sum of enthalpy of formation of products- sum of enthalpy of formation of reactants = (-2*46)- (-712 -3*436) = 1928 kJ/mol
ΔH = sum of bond energies of reactants- sum of bond energies of products
1928 = (3*436+712) – (3*x)
we can calculate the x (Avg bond enegy of N-H) from above
3x = (1308 + 712) – 1928
x = 92/3
x = 30.67 kJ/mol
Thanks and Regards

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