Question

The combustion of 1.760 g of propanol (C3H7OH) increases the temperature of a bomb calorimeter from 298.00K to 302.15K. The heat capacity of the bomb calorimeter is 14.24 kJ/K. Determine ΔH for the combustion of propanol to carbon dioxide gas and liquid water in kJ/mol.

Answer 1

Answer: The enthalpy of the reaction is -1709.76 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 = 14.24 kJ/K

$\Delta T$ = change in temperature = $T_2-T_1=(302.15-298)K=4.15K$

Putting values in above equation, we get:

$q=14.24kJ/K\times 4.15K=59.096kJ$

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 propanol = 1.760 g

Molar mass of propanol = 60 g/mol

Putting values in above equation, we get:

$\text{Moles of propanol}=\frac{1.760g}{60g/mol}=0.0293mol$

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

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

where,

q = amount of heat released = -50.096 kJ

n = number of moles = 0.0293 moles

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

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-50.096kJ}{0.0293mol}=-1709.76kJ/mol$

Hence, the enthalpy of the reaction is -1709.76 kJ

Answer 2

The ΔH for the combustion of propanol to carbon dioxide gas and liquid water in kJ/mol is - 1709.76

What is combustion?

Combustion reactions produce heat when reacts, the reaction involves oxygen

By the formula of

$Q = m \Delta Ct$

The temperature will be minus

Putting the values in the equation

$Q = 14.24 \times 4. 15 K = 59. 096$

Calculate the moles of propanol

$\rm Number\;of \;moles= \dfrac{1.760}{60} = 0.293$

Calculating the enthalpy

$\Delta H = rxn = \dfrac{50.096 }{0.293\;mol} = -1709.76$

Thus, the ΔH for the combustion of propanol to carbon dioxide gas and liquid water in kJ/mol is - 1709.76

Learn more about combustion

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