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

To solve this problem, we will use the equation:

ΔH = q/n

where ΔH is the molar enthalpy of combustion of isooctane, q is the heat absorbed by the calorimeter (calculated from the temperature increase and the heat capacity of the calorimeter), and n is the number of moles of isooctane that reacted, which can be calculated from the given mass and molar mass.

First, let's calculate the number of moles of isooctane:

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

Next, let's calculate the heat absorbed by the calorimeter:

q = CΔT = 5.46 kJ/°C x 10.0°C = 54.6 kJ

Finally, let's calculate the molar enthalpy of combustion of isooctane:

ΔH = q/n = 54.6 kJ / 0.01 mol = 5,460 kJ/mol

The enthalpy change is positive because the reaction is exothermic (it releases heat), so the answer is (C) -5,460 kJ/mol.

Isooctane Is a Good Model Compound for Gasoline