Calculate the heat of combustion for propane gas, C3H8(g), if 12.5 grams of propane are completely burned in excess oxygen gas and the products of reaction are cooled from 700°C to 25°C. The heat capacities of the reactants and products are as follows: C3H8(g): cp = 44.10 J/K O2(g): cp = 29.38 J/K CO2(g): cp = 36.8 J/K H2O(g): cp = 33.6 J/K Assume that the heat capacities of the products and reactants are constant over the temperature range of the reaction and that the heat lost to the surroundings is negligible.

Question

Calculate the heat of combustion for propane gas, C3H8(g), if 12.5 grams of propane are completely burned in excess oxygen gas and the products of reaction are cooled from 700°C to 25°C. The heat capacities of the reactants and products are as follows: C3H8(g): cp = 44.10 J/K

Answer 1

First, let's write the balanced combustion reaction for propane:C3H8 g  + 5O2 g  -> 3CO2 g  + 4H2O g Now, we need to determine the moles of propane burned:Moles of propane = mass / molar massMoles of propane = 12.5 g /  3 * 12.01 g/mol + 8 * 1.01 g/mol  = 12.5 g / 44.09 g/mol = 0.2834 molNext, we will calculate the moles of products formed:Moles of CO2 = 3 * moles of propane = 3 * 0.2834 mol = 0.8502 molMoles of H2O = 4 * moles of propane = 4 * 0.2834 mol = 1.1336 molNow, we will calculate the heat change for cooling the products from 700C to 25C:H_CO2 = moles of CO2 * cp_CO2 * T = 0.8502 mol * 36.8 J/mol*K *  25 - 700  = -0.8502 * 36.8 * -675 = 21256.9 JH_H2O = moles of H2O * cp_H2O * T = 1.1336 mol * 33.6 J/mol*K *  25 - 700  = -1.1336 * 33.6 * -675 = 25687.2 JThe total heat change for cooling the products is the sum of the heat changes for CO2 and H2O:H_total = H_CO2 + H_H2O = 21256.9 J + 25687.2 J = 46944.1 JNow, we need to calculate the heat of combustion per mole of propane. We can do this by dividing the total heat change by the moles of propane:H_combustion = H_total / moles of propane = 46944.1 J / 0.2834 mol = 165,600 J/molThe heat of combustion for propane gas is approximately 165,600 J/mol.