Question

The following information is given for benzene, C6H6, at 1atm: boiling point = 80.1 °C Hvap(80.1 °C) = 30.7 kJ/mol specific heat liquid = 1.74 J/g°C At a pressure of 1 atm, what is H in kJ for the process of condensing a 24.8 g sample of gaseous benzene at its normal boiling point of 80.1 °C. kJ

Answer 1

Answer: The heat required for the process is 4.24 kJ

Explanation:

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

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

Given mass of benzene = 24.8 g

Molar mass of benzene = 78.11 g/mol

Putting values in above equation, we get:

$\text{Moles of benzene}=\frac{24.8g}{78.11g/mol}=0.318mol$

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

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

where,

$q$ = amount of heat absorbed = ?

n = number of moles = 0.318 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction  = 30.7 kJ/mol

Putting values in above equation, we get:

$30.7kJ/mol=\frac{q}{0.318mol}\\\\q=(30.7kJ/mol\times 0.318mol)=4.24kJ$

Hence, the heat required for the process is 4.24 kJ