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InterviewSolution
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Bond dissociation enthalpy of `H_(2)` , `Cl_(2)` and `HCl` are `434, 242` and `431KJmol^(-1)` respectively. Enthalpy of formation of `HCl` isA. `93 KJ "mol"^(-1)`B. `-245 KJ "mol"^(-1)`C. `-93 KJ "mol"^(-1)`D. `245 KJ "mol"^(-1)` |
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Answer» Correct Answer - C Given, `DeltaH_(H-H)=434` KJ/mol `DeltaH_(Cl-Cl)=242` KJ/mol `DeltaH_(H-Cl) = 431` KJ/mol `(1)/(2)H_(2) + (1)/(2) Cl_(2) to HCl, DeltaH_(r)=?` `DeltaH_(r)=(1)/(2)xxDeltaH_(H-H)+(1)/(2)xxDeltaH_(Cl-Cl)-DeltaH_(H-Cl)` `=(1)/(2)xx434 + (1)/(2)xx 242 - 431` `=217xx121 - 431 =- 93` KJ/mol |
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