Based on the values of B.E. given, ΔHf° of N2H4(g) is :-Given : N – N = 159 kJ mol–1 H – H = 436 kJ mol–1 N ≡ N = 941 kJ mol–1 N – H = 398 kJ mol–1711 kJ mol–162 kJ mol–1–98 kJ mol–1–711 kJ mol–1
Question
Based on the values of B.E. given, ΔHf° of N2H4(g) is :-Given : N – N = 159 kJ mol–1 H – H = 436 kJ mol–1 N ≡ N = 941 kJ mol–1 N – H = 398 kJ mol–1711 kJ mol–162 kJ mol–1–98 kJ mol–1–711 kJ mol–1
Solution
Based on the given values of B.E., we can calculate the ΔHf° of N2H4(g) step by step:
- N – N = 159 kJ mol–1
- H – H = 436 kJ mol–1
- N ≡ N = 941 kJ mol–1
- N – H = 398 kJ mol–1
To calculate the ΔHf° of N2H4(g), we need to consider the formation reaction of N2H4(g) from its constituent elements:
N2(g) + 4H(g) → N2H4(g)
Using the given bond energies, we can calculate the energy change for breaking the bonds in the reactants and forming the bonds in the product:
Energy change = (4 × N – H) - (1 × N ≡ N + 4 × H – H)
Substituting the given values:
Energy change = (4 × 398) - (1 × 941 + 4 × 436)
Energy change = 1592 - (941 + 1744)
Energy change = 1592 - 2685
Energy change = -1093 kJ mol–1
Therefore, the ΔHf° of N2H4(g) is -1093 kJ mol–1.
Similar Questions
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