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InterviewSolution
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The standard enthalpy of formation of `NH_(3)` is `-46.0KJ mol^(-1)` . If the enthalpy of formation of `H_(2)` from its atoms is `-436KJ mol^(-1)` and that of `N_(2)` is `-712KJ mol^(-1)` , the average bond enthalpy of `N-H` bond in `NH_(3)` isA. `-964KJmol^(-1)`B. `+352KJmol^(-1)`C. `+105KJmol^(-1)`D. `-1102KJmol^(-1)` |
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Answer» Correct Answer - C `(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)rarrNH_(3)(g),Delta_(f)H=-46KJmol^(-1)` Bond enthalpy of `H_(2)=436KJmol^(-1)` [`.^(+)ve` sing is taken because energy is supplied to break the `H-H` bond into its atoms] Similary, bond enthalpy of `N_(2)=+712KJmol^(-1)` `Delta_(f)H=[(1)/(2)BE(N_(2))+(3)/(2)BE(H_(2))]-3BE(N-H)` `-46=[(1)/(2)xx712+(3)/(2)xx436]-3BE(N-H)` `3BE(N-H)=(1010)+46` `3BE(N-H)=1056` `BE(N-H)=1056//3=352KJmol^(-1)` |
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