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In a constant volume calorimeter 5.0 of a gas with molecular wt 40 was burnt in excess of oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 to 298.75K .due to combustion process. Given that the heat capacity of the calorimeter is2.5kJ /K the numerical value for the entalpy of combustion of the gas in kJ/mol

In a constant volume calorimeter 5.0 of a gas with molecular wt 40 was burnt in excess of oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 to 298.75K .due to combustion process. Given that the heat capacity of the calorimeter is2.5kJ /K the numerical value for the entalpy of combustion of the gas in kJ/mol

Grade:12th pass

2 Answers

souradeep
27 Points
6 years ago
number of moles(n )of the gas is=5/40=0.125
total enthalpy change is=heat capacity*change in temp=2.5*(298.75-298)=1.875
enthalpy of combustion per mole=1.875/0.125=15
Mushtaq Ahamad
15 Points
5 years ago
We know that,
For ideal gas under any process,
Change in enthalapy=molar heat capacity*change in temprature/no.of moles
Therefore,
Change in enthalapy=2.5*0.75/0.125
=1.875/0.125
=15kJ/mol

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