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For combustion of 1 mole of benzene at `25^(@)C`, the heat of reaction at constant pressure is `-780.9" kcal."` What will be the heat of reaction at constant volume ? `C_(6)H_(6(l))+7(1)/(2)O_(2(g))rarr 6CO_(2(g))+3H_(2)O_((l))`A. `-781.8" kcal"`B. `-"780.0 kcal"`C. `+"781.8 kcal"`D. `+"780.0 kcal"` |
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Answer» Correct Answer - B `q_(p)=q_(v)+Deltan_(g)RT, q_(v)=q_(p)-Deltan_(g)RT(DeltaH=DeltaU+Deltan_(g)RT)` `q_(v)=-780.9-(-1.5xx2xx298xx10^(-3))=-"780 kcal"` |
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