InterviewSolution
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InterviewSolution
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given `H_(2)O(l) to H^(+)(aq) + OH^(-)(aq),DeltaH= 57.32 kJ` `H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l),DeltaH =-286.02 kJ` then , calculate the enthalpy of formation of `OH^(-) at 25^(@)C`A. `-228.8kJ`B. `-343.52 kJ`C. `+228.8 kJ`D. `+ 343. 52 kJ` |
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Answer» Correct Answer - a consider the formation of `H_(2)O` `H_(2)(g)+1/2O_(2)(g) to H_(2)O(l),DeltaH=-286.20 kJ` `DeltaH_(r)= DeltaH_(t)[H_(2)O(l)]-DeltaH_(t)[H_(2)(g))]-1/2 DeltaH_(t)[O_(2)(g)] ` ` DeltaH_(t)(H_(2)O(l))=-286.20` now consdier the ionisation of `H_(2)O` `H_(2)O(l) to H^(+)(aq)+ OH^(-)(aq), DeltaH = 57.32 KJ H_(2)O(l) to H^(+)(aq)+ OH^(-)(aq), DeltaH = 57.32 KJ` |
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