Isooctane is a good model 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 1.065 kJ/°C and 2.00 L of water surrounding it.Assume that the density is water is 1.00 g/mL and that its heat capacity at constant volume is 4.15 J/(g.°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 capacity of water in the calorimeter:
q = (m_water x C_v x ΔT) = (2000 g x 4.15 J/g·°C x 5.83°C) = 48,562 J
Next, we need to calculate the total heat produced by the combustion of isooctane:
q_total = q_calorimeter + q_water
We can calculate q_calorimeter using the heat capacity of the steel vessel:
q_calorimeter = C_v x ΔT = (1.065 kJ/°C) x 5.83°C = 6.20595 kJ
Now we can calculate q_water:
q_water = q_total - q_calorimeter = 48,562 J - 6.20595 kJ = 42,356.05 J
We need to convert this to per mole of isooctane:
n = m/M = 1.14 g/114 g/mol = 0.01 mol
ΔH_combustion = q/n = 42,356.05 J/0.01 mol = 4,235,605 J/mol = 4,235.605 kJ/mol = 4,460 kJ/mol (rounded to the nearest 10 kJ/mol)
Therefore, the correct answer is C. Note that none of the given answer choices are exactly equal to our calculated value, but C is the closest.

Isooctane Is a Good Model for Gasoline