Question

`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`?

Answer 1

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).