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H_(2)O(l)hArrH^(+)(aq)+OH^(-)(aq),DeltaH=+13.7 kcal/mol K_(eq) at 25^(@)C=10^(-14). Calculate K_(eq) at 60^(@)C? |
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Answer» Solution :`LOG(K_(2))/(K_(1))=(DeltaH)/(2.303R)[1/(T_(1))-1/(T_(2))]` `log(K_(2))/(10^(-14))=(13.7)/(2.303xx2xx10^(-3))[1/298-1/333]` `K_(2)=1.26xx10^(-12)` |
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