Question

When 0.0901 mol of an unknown hydrocarbon is burned in a bomb calorimeter, the calorimeter increases in temperature by 2.19°C. If the heat capacity of the bomb calorimeter is 1.229 kJ/°C, what is the heat of combustion for the unknown hydrocarbon?

Answer 1

The heat of combustion for the unknown hydrocarbon is -29.9 kJ/mol

Step-by-step explanation:

Step 1: Data given

Number of moles of the unknown hydrocarbon = 0.0901 moles

The temperature in the calorimeter rises with 2.19 °C

The heat capacity of the calorimeter is 1.229 kJ/°C

Step 2: Calculate heat

Q = c(calorimeter) * ΔT

⇒with Q = the heat transfer = TO BE DETERMINED

⇒with c = the heat capacity of the calorimeter = 1.229 kJ/°C

⇒with ΔT = The change in temperature = 2.19 °C

Q = 1.229 kJ/°C * 2.19 °C

Q = 2.6915 kJ

Step 3: Calculate the heat of combustion for the unknown hydrocarbon

Heat of combustion = Q/moles

Heat of combustion = 2.6915 kJ / 0.0901 moles

Heat of combustion = 29.9 kJ/mol ( the sign should be negative )

Heat of combustion = -29.9 kJ/mol

The heat of combustion for the unknown hydrocarbon is -29.9 kJ/mol

Answer 2

The heat of combustion for the unknown hydrocarbon is -29.87 kJ/mol

Step-by-step explanation:

Heat capacity of the bomb calorimeter = C = 1.229 kJ/°C

Change in temperature of the bomb calorimeter = ΔT = 2.19°C

Heat absorbed by bomb calorimeter = Q

`Q=C* \Delta T`

`Q=1.229 kJ/^oC* 2.19^oC=2,692 kJ`

Moles of hydrocarbon burned in calorimeter = 0.0901 mol

Heat released on combustion = Q' = -Q = -2,692 kJ

The heat of combustion for the unknown hydrocarbon :

`(Q')/(0.090 mol)=(-2,692 kJ)/(0.0901 mol)=-29.87 kJ/mol`