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

Answer:

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

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

$Q=C\times \Delta T$

$Q=1.229 kJ/^oC\times 2.19^oC=2,692 kJ$

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 :

$\frac{Q'}{0.090 mol}=\frac{-2,692 kJ}{0.0901 mol}=-29.87 kJ/mol$

Answer 2

Answer:

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

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