In a constant volume calorimeter, $$3.5$$ g of a gas with molecular weight $$28$$ was burnt in excess oxygen at $$298.0$$ K. The temperature of the calorimeter was found to increase from $$298.0$$ K to $$298.45$$ K due to the combustion process. Given that the heat capacity of the calorimeter is $$2.5kJK-$$ the numerical value for the enthalpy of combustion of the gas in $$kJmol-$$. is:

Question

In a constant volume calorimeter, $$3.5$$ g of a gas with molecular weight $$28$$ was burnt in excess oxygen at $$298.0$$ K. The temperature of the calorimeter was found to increase from $$298.0$$ K to $$298.45$$ K due to the combustion process. Given that the heat capacity of the calorimeter is $$2.5kJK-$$ the numerical value for the enthalpy of combustion of the gas in  $$kJmol-$$. is:

Infinity Learn · Accepted Answer

Since the combustion takes place at constant volume, the energy released ΔU will be given as:Δ$$U=Z$$×ΔT×Mm                                  ... (1)Where$$          $$Z=$$ heat capacity of the calorimeter $$=2.5kJK$$−Δ$$T=$$ change in temperature $$=298.45$$−$$298.0=0.45KM=$$ molar mass of gas $$=28g$$ mol −m $$=$$ mass of the substance $$=5.3gSubstituting$$ the values in $$(1), we get :Δ$$U=2.5kJK$$−×$$0.45K$$×$$28gmol$$−$$5.3g=9.0kJmol$$−.

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