In a constant volume calorimeter, 3.5 g of a gas with moleular 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 2.5 KJK^(-1), the numerical value for the enthalphy of combustion of the gas in KJ mol^(-1) is .

In a constant volume calorimeter, 3.5 g of a gas with moleular weight

1.	In a constant volume calorimeter, 3.5 g of a gas with moleular 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 2.5 KJK^(-1), the numerical value for the enthalphy of combustion of the gas in KJ mol^(-1) is .
Answer» Solution :`N=(3.5)/(28)` `DeltaT = T_(2)- T_(1) = 298.45 - 298 = 0.45` `C_(V)=2.5KJ K^(-1) = 2500 JK^(-1)` `C_(p) = C_(v) + R = 2500 + 8.314 = 2508.314 JK^(-1)` `Q_(p) = C_(p)DeltaT = 1128.74 J` `DELTAH = (Q_(p))/(n) = (1128.74 )/(3.5//28) = 9030 J mol^(-1) = 9.030 KJ mol^(-1) = 9 KJ mol^(-1).`