The enthalpy of combustion of liquid octane, C8H18(l) to gaseous products, is -5090 kJ.mol-1. Use th — Thermodynamics Chemistry Question
Problem Context
The enthalpy of combustion of liquid octane, C8H18(l) to gaseous products, is -5090 kJ.mol-1. Use this value to answer the questions below, assuming a temperature of 100 ˚C.
Write a balanced equation for the complete combustion of liquid octane C8H18(l).
Model Answer
Write balanced equations to represent the processes responsible for K1 and K2.
2 C8H18 (l) + 25 O2 (g) ! 16 CO2 (g) + 18 H2O (g)
Determine the molar enthalpy of formation, ∆Hf̊, for liquid octane, C8H18(l). [∆Hf˚ kJ.mol-1; CO2(g) -393.5, H2O(g) -241.8]
Model Answer
∆Hrxn = 8∆Hf(CO2) + 9∆Hf(H2O) - ∆Hf(C8H18)
–5090 kJ = 8(–393.5 kJ) + 9(–241.8 kJ) - ∆Hf(C8H18)
∆Hf(C8H18) = –3148 – 2176 + 5090 kJ = -234.2 kJ
Calculate the value of the internal energy change, ∆E˚, for the combustion reaction.
Model Answer
∆H = ∆E + ∆nRT so ∆E = ∆H - ∆nRT
= –5090000 J – (4.5 mol)(8.314 J. mol-1 . K-1 )(373 K)
= –5090000 J – 13955 J
= –5104000 J = -5104 kJ /mol C8H18(l)
If ∆G˚ for the combustion is -5230 kJ.mol-1 of octane, calculate the value of ∆S˚. Comment on the sign of ∆S˚ relative to the equation written above.
Model Answer
∆G° = ∆H° − T∆S°
-5230 kJ . mol-1 = -5090 kJ . mol-1 – 373 K (∆So)
so ∆S° = (-5090 kJ mol-1 + 5230 kJ mol-1) / 373 K
= 0.375 kJ.mol-1.K-1
The increase in ∆S˚ is consistent with the formation of more moles of gas during the reaction.
State whether the heat associated with the combustion of liquid octane in a bomb calorimeter represents ∆H˚ or ∆E˚. Explain your reasoning.
Model Answer
Heat in a bomb calorimeter is ∆E˚ (q at constant volume) - no credit unless there is a discussion of zero work under constant volume.