Question

A bomb calorimeter has a heat capacity of 2.47 kJ/K. When a 0.105-g sample of ethylene (C2H4) was burned in this calorimeter, the temperature increased by 2.14 K. Calculate the energy of combustion for one mole of ethylene. A. –1.41 × 103 kJ/mol B. –660 kJ/mol C. –5.29 kJ/mol D. –0.259 kJ/mol E. –50.3 kJ/mol

Answer 1

A. –1.41 × 10³ kJ/mol

Step-by-step explanation:

According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the calorimeter is zero.

Qcal + Qcomb = 0

Qcomb = -Qcal [1]

We can calculate the heat absorbed by the calorimeter using the following equation:

Qcal = Ccal × ΔT

where,

Ccal is the heat capacity

ΔT is the change in the temperature

From [1]

Qcomb = -Qcal = -Ccal × ΔT = -2.47 kJ/K × 2.14 K = -5.29 kJ

We know that the molar mass of ethylene is 28.05 g/mol. Then, the heat released per mole of ethylene is:

`(-5.29kJ)/(0.105g) .(28.05g)/(mol) =-1.41 * 10^(3) kJ/mol`