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)

Answer 1

The question is incomplete, here is the complete question.

A sample of ethanol $(C_2H_5OH)$, weighing 6.83 g underwent combustion in a bomb calorimeter by the following reaction:

$C_2H_5OH(l)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(l)$

If the heat capacity of the calorimeter and contents was 18.1 kJ/°C and the temperature of the calorimeter rose from 25.50°C to 36.73°C. What is the $\Delta H$ of the reaction?

Answer: The enthalpy of the reaction is -1355.1 kJ

Explanation:

To calculate the heat absorbed by the calorimeter, we use the equation:

$q=c\Delta T$

where,

q = heat absorbed

c = heat capacity of calorimeter = 18.1 kJ/°C

$\Delta T$ = change in temperature = $T_2-T_1=(36.73-25.50)^oC=11.23^oC$

Putting values in above equation, we get:

$q=18.1kJ/^oC\times 11.23^oC=203.26kJ$

Heat absorbed by the calorimeter will be equal to the heat released by the reaction.

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of ethanol = 6.83 g

Molar mass of ethanol = 46 g/mol

Putting values in above equation, we get:

$\text{Moles of ethanol}=\frac{6.83g}{46g/mol}=0.15mol$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = -203.26 kJ

n = number of moles = 0.15 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-203.26kJ}{0.15mol}=-1355.1kJ/mol$

Hence, the enthalpy of the reaction is -1355.1 kJ