Question

13. Propane burns in air according to the following equation:

C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g) ∆H = -2043 kJ

If 5.00 g of propane are reacted at a constant pressure of 1.15 atm leading to a change in volume of +8.6 L, calculate ∆E and the work done (in kJ). (MM C3H8 = 44.09 g/mol).

Using the information given (∆Hrxn = -2043 kJ) and any needed values from you text, calculate the ∆H o f for propane (C3H8).

Answer 1

Ans:

C₃H₈ (g) + 5 O₂ (g) → 3 CO₂ (g) + 4 H₂O   ∆H = -2043 kJ

P = 1.15 atm, w = 5 gm, ∆V = 8.6 L, MW of C₃H_{8 =} 44.06 g/mol

1) As system is doing work on the surroundings,

W = - P∆V = - 1.15 atm X 8.6 L = - 9.89 X 0.101325 KJ

            Therefore, W = - 1.0021 KJ.

2) ΔH_reaction = q_{at constant pressure} = q_reaction

Hence, q = - 2043 KJ

∆E = q + W = -2043 + (-1.0021)

= - 2.044 X 10³ KJ.

3) ∆H_o = 5 X 10^-3 kg X 1000 g/1 kg X 1 mol/44.09 g X – 2043 KJ/1 mol

= - 2.316 X 10² KJ.