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For the reaction , `H_(2)(g)+1//2O_(2)(g)=H_(2)O(l), Delta C_(p)=7.63 "cal/deg" , Delta H_(25^(@)C)=68.3` Kcal, what will be the value (in Kcal) of `Delta H` at `100^(@)C` :A. `7.63xx(373-298)-68.3`B. `7.63xx10^(-3)(373-298)-68.3`C. `7.63xx10^(-3)(373-298)+68.3`D. `7.63xx(373-298)+68.3` |
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Answer» Correct Answer - C `(Delta H_(2)-Delta H_(1))/(T_(2)-T_(1))=Delta C_(p)` `Delta H_(2)=Delta C_(P)(T_(2)-T_(1))+Delta H_(1))` `Delta H_(2)=7.63 (100-25)+68.3` |
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