Based on the values of B.E. given, ∆ f H o of N 2 H 4 (g) is: Given: N–N = 159 kJ mol –1…
Chemistry · JEE Main · NTA Exams — Chemical Thermodynamics
Based on the values of B.E. given, ∆fHo of N2H4(g) is:
Given: N–N = 159 kJ mol–1; H–H = 436 kJ mol–1
N\(\equiv\)N = 941 kJ mol–1, N–H = 398 kJ mol–1
Given: N–N = 159 kJ mol–1; H–H = 436 kJ mol–1
N\(\equiv\)N = 941 kJ mol–1, N–H = 398 kJ mol–1
- 711 kJ mol–1
- 62 kJ mol–1
- –98 kJ mol–1
- –711 kJ mol–1
Answer
(B) 62 kJ mol –1
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