Notifications
Clear all
0
25/09/2021 4:12 pm
Topic starter
The standard enthalpy of formation of NH3 is -46.0 kJ/mol. If the enthalpy of formation of H2 from its atoms is -436 kJ/mol and that of N2 is -712 kJ/mol, the average bond enthalpy of N - H bond in NH3 is:
(a) -1102 kJ/mol
(b) -964 kJ/mol
(c) +352 kJ/mol
(d) +1056 kJ/mol
1 Answer
0
25/09/2021 4:22 pm
Correct answer: (b) -964 kJ/mol
Explanation:
Given \(\frac{1}{2}\)N2 + \(\frac{3}{2}\)H2 ⇌ NH3;
ΔHf = -46.0 kJ / mol
H + H ⇌ H2; ΔHf = -436 kJ/ mol
N + N ⇌ N2; ΔHf = -712 kJ/mol
ΔHf(NH3) = \(\frac{1}{2}\)ΔHN-N + \(\frac{3}{2}\)ΔHH-H - ΔHN-F
-46 = \(\frac{1}{2}\)(-712) + \(\frac{3}{2}\)(-436) - ΔHN-F
On calculation
ΔHN-F = -964 kJ/ mol