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Heat of reaction for `C_(6)H_(12)O_(6(s))+6O_(2(g)) rarr 6CO_(2(g))+6H_(2)O_(v)` at constant pressure is `-651 kcal` at `17^(@)C`. Calculate the heat of reaction at constant volume at `17^(@)C`. |
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Answer» `DeltaH = DeltaU +DeltanRT….(i)` Given `DeltaH =- 651 xx10^(3) cal, R = 2 cal` Substituting all value in equation (i), we get `T = 290 K`, `Deltan = 6+6 -6 = 6` `:. =- 651 xx 10^(3) = DeltaU +6 xx 2xx 290` and `DeltaU =- 654480 cla or -654.48 kcal ` Note: For `C_(6)H_(12)O_(6)(g) +6O_(2)(g) rarr 6CO_(2)(g) +6H_(2)O(v) Deltan =` Number of products molecules -Number of reactants molecules (only in gaseous phase) `=6 +6 -6 = 6` Here, `H_(2)O(v)` is also in gaseous state. |
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