Question

A sample of ethanol (C2H5OH), weighing 6.83 g underwent combustion in a bomb calorimeter by the following reaction:C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (l)If the heat capacity of the calorimeter and contents was 18.1 kJ / oC and the temperature of the calorimeter rose from 25.50 to 36.73, (1) what is the ΔH of the reaction?

Answer 1

Answer:

-1.37 × 10³ kJ/mol

Explanation:

Let's consider the combustion of ethanol.

C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l)

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

Qcomb + Qcal = 0

Qcomb = - Qcal [1]

The heat absorbed by the bomb calorimeter can be calculated using the following expression.

Qcal = Ccal . ΔT = 18.1 kJ/°C . (36.73°C - 25.50°C) = 203 kJ

where

Ccal: heat capacity of the calorimeter

ΔT: change in the temperature

From [1],

Qcomb = -203 kJ

The molar mass of ethanol is 46.07 g/mol. The enthalpy of the combustion is:

$\Delta H =\frac{-203kJ}{6.83g} .\frac{46.07g}{mol} =-1.37 \times 10^{3} kJ/mol$