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InterviewSolution
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The enthalpy of neutralisation of a strong acid by a string base is `-57.32 kJ mol^(-1)`. The enthalpy of formation of water is `-285.84 kJ mol^(-1)`. The enthalpy of formation of hydroxy`1` ion isA. `+228.52 kJ mol^(-1)`B. `-114.26 kJ mol^(-1)`C. `-228.52 kJ mol^(-1)`D. `+114.2 kJ mol^(-1)` |
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Answer» The process of neutralisation is `H^(o+)(aq) + overset(Theta)OH (aq) rarr H_(2)O(l),DeltaH^(Theta) =- 57.32 kJ mol^(-1)` `DeltaH_(reaction)^(Theta)=sum` Heat of formation of products `-sum`Heat of fromation of reactants `= Delta_(f)H_(H_(2)O(l))^(Theta)-(Delta_(f)H_(H^(o+)(aq))^(Theta) +Delta_(f)H_(OH_((aq))^(Theta))^(Theta))` `-57.32 =- 285.84 +57.32 =- 228.52 kJ` `x =- 285.84 + 57.32 =- 228.52 kJ` Thus, the entalpy of formation of hydroxy`1` ion is `-228.52 kJ`. |
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