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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 increases from 298.0 ~K to 298.45 ~K due to the combustion process. Given, that the heat capacity of the calorimeter is 2.5 kJ K ⁻¹ , the numerical value for the enthalpy of combustion of the gas in kJ mol ⁻¹ is

Correct answer

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Step-by-step solution

The temperature rise is : T=T₂-T₁=298.45-298=0.45 ~K This indicates that heat produced from combustion of 3.5 ~g of compound rises temperature of calorimeter by 0.45 ~K . Heat produced =0.45 ~K 2.5 k JK ⁻¹=1.125 ~kJ Heat produced from 28 ~g of compound (1.0 ~mol )= 1.125 3.5 28=9 ~kJ

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