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Bond dissociation enthalpies of `H_(2)(g)` and `N_(2)(g)` are `436.0 kJ mol^(-1)` and `941.8 kJ mol^(-1)`, respectively, and ethalpy of formation of `NH_(3)(g)` is `-46kJ mol^(-1)`. What is the enthalpy fi atomisation of `NH_(3)(g)`?. What is the avergae bond ethalpy of `N-H` bond? |
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Answer» `N_(2)(g) +3H_(2)(g) rarr 2NH_(3)(g),DeltaH =-2 xx 46 kJ mol^(-1)` `DeltaH = sum(BE)_(R) -sum(BE)_(P)` `=(941.8 +3 xx 436) - (6x) =- 2 xx 46` (Here `x = BE` of `N-H` bonds) `x = 390.3 kJ mol^(-1)` `NH_(3) rarr N +3(H)` Heat of atomisaiton `=3xx 390.3 = 1170.9 kJ mol^(-1)` |
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