1.

When `0.1025 g` of benzoic acid was burnt in a bomb calorimeter the temperature of the calorimeter increased by `2.165^(@)C`. For benzoic acid `DeltaH_("comb")^(@)= -3227 kJ mol^(-1)`. Calculate the heat capacity of the calorimeter.

Answer» The equation for the combustion is ,
`C_(7)H_(6)O_(2)(s) + 7.5 O_(2)(g) rarr 7CO_(2)(g) + 3H_(2)O(l), " " Delta H^(@) = 3227 kJ`
Since `7.5` moles of `O_(2)` gas is needed, and 7 moles of `CO_(2)` is produced, some pressure-volume work is done, to the calorimeter:
`PV = Delta n_(g) R T`, where `Delta n =(7-7.5) = -0.5 mol`
`Delta E = Delta H - Delta Hn_(g) RT`
`= -3227 - (0.5 xx 8.314298 xx 298)`
`= -3226kJ//mol` (a small correction)
The amount of heat produced by 0.1025 g benzoic acid is
`q = 0.1025//122.13 xx 3226 = 2.680 kJ`
Thus, the heat capacity is
`C = q_(v)//Delta T = 2.680//2.165 = 1.238 kJ//K`.
After the heat capacity is determined, the calorimeter is ready to be used to measure the enthalpy of combustion of other substances.


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