InterviewSolution
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InterviewSolution
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The enthalpy of reaction for the reaction `:` `2H_(2)(g) + O_(2)(g) rarr 2H_(2)O(l) ` is `Delta_(r)H^(c-) =- 572 k J mol^(-1)` What will be standard enthalpy of formation of `H_(2)O(l) ` ? |
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Answer» Given`: 2H_(2)(g) + O_(2)(g) rarr2H_(2)O(l), Delta_(r)H^(@)= - 572kJ mol^(-1)` Enthalpy of formation is the enthalpy change of the reaction when 1 mole of the compound is formed from its elements, i.e., we aim at `H_(2)(g)+ (1)/(2) O_(2)(g) rarr H_(2)O(l), Delta_(r)H^(@) = ?` This can be obtained by dividing the given equation by 2. Hence, `Delta_(2)H^(@) ( H_(2)O) = ( - 572kJ mol^(-1))/( 2) =-286kJ mol^(-1)` |
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