Propane is often used as a home fuel in areas where natural gas is not available

Question

Propane is often used as a home fuel in areas where natural gas is not available. When 3.00 grams of propane (C_3 H_8 (g)) are burned in excess oxygen in a bomb calorimeter that has a total heat capacity of 32.7 kJ/K. the temperature of the calorimeter increases by 4.25 K. a. Calculate the heat of the calorimeter, q_cal. b. Calculate, by using the First law of Thermodynamics, the heat of the reacton, q_rxn. c. Assuming that delta E = delta H, calculate the standard enthalpy of combustion in (i) kilojoules per gram of propane, and (ii) kilojoules per mole of propane.

Explanation / Answer

Heat of calorimeter= heat capacity* temperature difference= 32.7*4.25 Kj=139 KJ

For adiabatic system, heat of calorimeter= heat of reaction =139 KJ

This is heat liberated from 3 gm of propane. Molar mass of propane= 44

Moles of propane= mass/molar mass= 3/44=0.068

Heat of reaction= 139/0.068 Kj/moles=2044 Kj/mole

deltaH= 2044 Kj/mole

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.