A 1.800 g sample of octane, C8H18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/°C. The temperature of the calorimeter plus contents increased from 21.12°C to 29.85°C.
What is the heat of combustion per gram of octane?
What is the heat of combustion per mole of octane?
heat released = q = Ccal*delta T = 11.66 x (Tfinal-Tinitial) in kJ.
q/1.800 = heat/gram in kJ/gram
(q/1.800) x molar mass Octane = delta H/mol in kJ/mol
0.495
To calculate the heat of combustion per gram of octane, we can use the equation:
Heat of combustion per gram = Total heat transferred / Mass of octane
Given data:
Mass of octane = 1.800 g
Total heat capacity of the calorimeter = 11.66 kJ/°C
Change in temperature of the calorimeter plus contents = 29.85°C - 21.12°C = 8.73°C
First, we need to calculate the total heat transferred during the combustion:
Total heat transferred = Total heat capacity × Change in temperature
Total heat transferred = 11.66 kJ/°C × 8.73°C
Total heat transferred = 101.7878 kJ
Now, we can calculate the heat of combustion per gram of octane:
Heat of combustion per gram = Total heat transferred / Mass of octane
Heat of combustion per gram = 101.7878 kJ / 1.800 g
Heat of combustion per gram of octane ≈ 56.55 kJ/g
To calculate the heat of combustion per mole of octane, we need to convert grams to moles:
Molar mass of octane (C8H18) = 114.22 g/mol
Number of moles of octane = Mass of octane / Molar mass of octane
Number of moles of octane = 1.800 g / 114.22 g/mol
Number of moles of octane ≈ 0.01574 mol
Now, we can calculate the heat of combustion per mole of octane:
Heat of combustion per mole = Total heat transferred / Number of moles of octane
Heat of combustion per mole = 101.7878 kJ / 0.01574 mol
Heat of combustion per mole of octane ≈ 6478.6 kJ/mol
To determine the heat of combustion per gram of octane, you need to first calculate the heat transferred to the calorimeter during the combustion process. Then, divide this value by the mass of the octane sample.
The equation for heat transfer in this case can be expressed as:
q = (m × c × ΔT)
Where:
q = heat transferred (in joules)
m = mass of the sample (in grams)
c = specific heat capacity of the calorimeter (in J/g·°C)
ΔT = change in temperature (in °C)
We need to convert the mass of the octane sample from grams to kilograms, which is 0.001800 kg. The specific heat capacity of the calorimeter is given as 11.66 kJ/°C, which is 11.66 J/g·°C.
Substituting the values into the equation, we get:
q = (0.001800 kg × 11.66 J/g·°C × (29.85 - 21.12) °C)
Simplifying this:
q = 0.0225876 kJ = 22.5876 J
The next step is to divide this heat transferred by the mass of the octane sample in grams:
Heat of combustion per gram of octane = q / m
= 22.5876 J / 1.800 g
Calculating this:
Heat of combustion per gram of octane ≈ 12.55 J/g
To determine the heat of combustion per mole of octane, we need to calculate the number of moles of octane burned. This can be done using the molar mass of octane, which is 114.22 g/mol.
First, divide the mass of the octane sample by its molar mass to obtain the number of moles:
Number of moles of octane = mass of octane / molar mass of octane
= 1.800 g / 114.22 g/mol
Calculating this:
Number of moles of octane ≈ 0.015752 mol
Now, we can determine the heat of combustion per mole of octane:
Heat of combustion per mole of octane = Heat of combustion per gram of octane × molar mass of octane
= 12.55 J/g × 114.22 g/mol
Calculating this:
Heat of combustion per mole of octane ≈ 1432 J/mol