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 $\circ$ C. The calorimeter consisted of two components: a steel vessel with a heat capacity of 1.00 kJ/ $\circ$ 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 . $\circ$ C), respectively. Use these data to determine the molar enthalpy of combustion of isooctane.

Question

Quizplus · Accepted Answer

First, we need to calculate the heat absorbed by the water in the calorimeter:

q = m × c × ΔT

where q is the heat absorbed, m is the mass of water, c is the specific heat capacity of water, and ΔT is the temperature increase.

m = V × ρ = 2.00 L × 1.00 g/mL = 2000 g

q = 2000 g × 4.184 J/(g·°C) × 5.83°C = 48,855.04 J

Next, we need to calculate the total heat released by the combustion of 1 mole of isooctane:

q = ΔH × n

where q is the heat released, ΔH is the molar enthalpy of combustion, and n is the number of moles of isooctane.

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

q = (1.00 kJ/°C + 48,855.04 J)/(0.01 mol) = 4,885,604 J/mol = 4885.6 kJ/mol

Finally, we need to correct for the heat absorbed by the steel vessel:

q = C × ΔT

where q is the heat absorbed, C is the heat capacity of the steel vessel, and ΔT is the temperature increase.

q = 1.00 kJ/°C × 5.83°C = 5.83 kJ

The total heat released is therefore:

q total = q isooctane - q steel + q water = 4885.6 kJ/mol - 5.83 kJ + 48.86 kJ = 4885.6 kJ/mol + 43.03 kJ = 4928.6 kJ/mol

Therefore, the best choice is (C) (49,460 kJ/mol), which is closest to our calculated value of 4928.6 kJ/mol.

Isooctane Is a Good Model Compound for Gasoline ∘\circ∘ C the Calorimeter Consisted of Two Components: a Steel Vessel

Isooctane Is a Good Model Compound for Gasoline $\circ$ C the Calorimeter Consisted of Two Components: a Steel Vessel