The standard enthalpy of formation of `NH_(3)` is `-46.0kJmol^(-1)`. If the enthalpy of formation of `H_(2)` from its atoms is `-436kJmol^(-1)`, the a
The standard enthalpy of formation of `NH_(3)` is `-46.0kJmol^(-1)`. If the enthalpy of formation of `H_(2)` from its atoms is `-436kJmol^(-1)`, the average bond enthalpy of N-H bond in `NH_(3)` is
A. `-964kJmol^(-1)`
B. `-352kJmol^(-1)`
C. `+1056kJmol^(-1)`
D. `-1102kJmol^(-1)`
4 views
1 Answers
Correct Answer - B
Given, `(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) to NH_(3)(g),DeltaH_(f)^(@)=-46.0" kJ "mol^(-1)`
`2H(g)toH_(2)(g),DeltaH_(f)^(@)=-436" kJ "Mol^(-1)`
`2N(g)toN_(2)(g),DeltaH_(f)^(@)=-712" kJ "mol^(-1)`
Bond enthalpy of `NH_(3)`
`=3BE_(N-H)[(1)/(2)BE_(N-N)+(3)/(2)BE_(H-H)]`
`-46=3N-N-[-(1)/(2)xx712-(3)/(2)xx436]`
`3N-H=-1056" kJ "mol^(-1)`
`N-H=(1056)/(3)=-352" kJ "mol^(-1)`
4 views
Answered