0.4 g of Propane burns completely at 300 K in a bomb calorimeter. The temperature of the calorimeter and surrounding water rises by 0.4^0 C . If the heat capacity of the calorimeter and contents is 24 ~kJ ~K ⁻¹ what is the Enthalpy for the reaction? (Assume that propane gas shows ideal behaviour).

Step-by-step solution

The combustion reaction of propane is given by: C₃H₈(g) + 5O₂(g) 3CO₂(g) + 4H₂O(l) . The heat released in the bomb calorimeter is q_ v = C T , where C = 24 kJ K ⁻¹ and T = 0.4 K . q_ v = 24 0.4 = 9.6 kJ . The number of moles of propane ( C₃H₈ ) is n = 0.4 g 44 g mol ⁻¹ = 0.4 44 mol = 1 110 mol . The internal energy change for the combustion of 1 mole of propane is U = - q_ v n = -9.6 110 = -1056 kJ mol ⁻¹ . The enthalpy change is given by H = U + n_ g RT , where n_ g = (3 - 1 - 5) = -3 . Using R = 8.314 10⁻³ kJ K ⁻