Isooctane is a good model compound for gasoline. When 1.14 g of isooctane (114 g/mol) reacts with excess oxygen in a constant pressure calorimeter, the temperature of the calorimeter increases by 10.0^oC. The heat capacity of the calorimeter is 5.46 kJ/^oC. Determine the molar enthalpy of combustion of isooctane.

Question

Quizplus · Accepted Answer

The molar enthalpy of combustion (∆Hcomb) can be calculated using the formula:

∆Hcomb = -qrxn/n

where qrxn is the heat released or absorbed during the reaction and n is the number of moles of the substance being combusted.

From the given information, we have:

n = 1.14 g / 114 g/mol = 0.01 mol

qrxn = -Cp∆T = -(5.46 kJ/°C)(10.0°C) = -54.6 kJ

Therefore, ∆Hcomb = -qrxn/n = -(-54.6 kJ)/0.01 mol = 5460 kJ/mol

The answer is (C) (𝛥Hcomb = -5460 kJ/mol).

Isooctane Is a Good Model Compound for Gasoline