The enthalpy changes at 25°C in successive breaking of O–H bonds of water are : H 2 O(g)…
Chemistry · JEE Main · NTA Exams — Chemical Thermodynamics
The enthalpy changes at 25°C in successive breaking of O–H bonds of water are :
H2O(g) \(\text { ⟶ }\) H(g) + OH(g) ∆H = 498 kJ mol–1
OH(g) \(\text { ⟶ }\) H(g) + O(g) ∆H = 428 kJ mol–1
the bond enthalpy of the O–H bond is :
H2O(g) \(\text { ⟶ }\) H(g) + OH(g) ∆H = 498 kJ mol–1
OH(g) \(\text { ⟶ }\) H(g) + O(g) ∆H = 428 kJ mol–1
the bond enthalpy of the O–H bond is :
- 498 kJ mol–1
- 463 kJ mol–1
- 428 kJ mol–1
- 70 kJ mol–1
Answer
(B) 463 kJ mol –1
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