Isooctane is a good model compound for gasoline. When 1.14 g of isooctane
(C₈H₁₈, molar mass = 114 g/mol) reacted with excess oxygen in a constant volume calorimeter, the temperature of the calorimeter increased by 5.83°C. The calorimeter consisted of two components: a steel vessel with a heat capacity of 1.00 kJ/°C and 2.00 L of water surrounding it. The density and specific heat capacity of water are 1.00 g/mL and 4.184 J/(g . °C) , respectively. Use these data to determine the molar enthalpy of combustion of isooctane.

Question

Quizplus · Accepted Answer

First, calculate the heat capacity of the water (assuming it's at room temperature, 25°C):

qwater = mcΔT = (2000 g) (4.184 J/g°C) (5.83°C) = 48,824 J

Next, calculate the heat released by the combustion reaction using the heat capacity of the calorimeter and the temperature change:

qrxn = -qcal = -CcalΔT = -(1000 g°C) (5.83°C) = -5830 J

Since the reaction is taking place at constant volume, the heat released by the combustion reaction is equal to the internal energy change (ΔU) of the system. We can relate this to the molar enthalpy of combustion (ΔHc) using the following equation:

ΔHc = -ΔU/n

where n is the number of moles of isooctane. To calculate n, we need to use the mass of isooctane and its molar mass:

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

Plugging in all the values, we get:

ΔHc = -(-5830 J/0.01 mol) = 583,000 J/mol = -5,460 kJ/mol

Therefore, the best choice is C, -5,460 kJ.

Isooctane Is a Good Model Compound for Gasoline