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`0.562g` of graphite kept in a bomb calorimeter in excess of oxygen at `298K` and 1 atmospheric pressure was burnt according to the equation, `C_(Graph ite)+O_(2(g))rarr CO_(2(g))` durgin the reaction, temperature rises from `298K` o `298.89K`. If the heat capacity of the calorimeter and its contents is `20.7 kJ//K`, what is the enthalpy change for the above reaction at `298K` and `1 atm`? |
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Answer» Suppose `q` is the quantity of heat from the reaction mixture and `C_(V)` is the capacity of the calorimeter and the content, then the quantity of heat absorbed by the calorimeter`=C_(V)xxDeltaT` Quantity of heat from the reaction will have the same magnitude but opposite sign, i.e., `q=C_(V)xxDeltaT=-20.7 kJ K^(-1)xx(298.89-298)K=-18.4 kJ` (Here, negative sign indicates the exothermic nature of the reaction) Thus, `DeltaH` for the combustion of the `0.562 g` of carbon `=-18.4 kJ` For combustion of `1 mol` of carbon, heat released `=1 mol Cxx(12.0 g C)/(1 mol C)xx((-18.4 kJ))/(0.562 g C)` `=- 3.9xx10^(2) kJ` (result reported to two significant figures) Thus, enthalpy for combustion of graphite `=-3.9xx10^(2) kJ mol^(-1)` Enthalpy change, `DeltaH` of a reaction Reachants `rarr` Products is represneted by `Delta_(r)H=("sum of enthalpies of products")- ("sum of enthalpies of reactants")` `=SigmaH_("Products")-SigmaH_("reactants")` (Here symbol `Sigma("sigma")` is used for summation). |
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