Question

The fluorocarbon compound C2Cl3F3 has a normal boiling point of 47.6 °C. The specific heats of C2Cl3F3(l) and C2Cl3F3(g) are 0.91 J/g.K and 0.67 J/g.K, respectively. The heat of vaporization for the compound is 27.49 kJ/mol.
Calculate the heat required to convert 75.0 g of C2Cl3F3 from a liquid at 11.50 °C to a gas at 80.30 °C.

Answer 1

Answer:

The heat required to convert 75.0 g of $C_2Cl_3F_3$ from a liquid at 11.50 °C to a gas at 80.30 °C is 15.103 kiloJoules.

Explanation:

1) Heat required to raise the temperature of fluorocarbon compound from 11.50°C to 47.6 °C= $Q_1$

$C_2Cl_3F_3(l,284.65 K)\rightarrow C_2Cl_3F_3(l,320.75 K)$

Mass of fluorocarbon compound = m = 75.0 g

Specific heat of fluorocarbon compound ,c= 0.91 J/gK

Change in temperature = $\Delta T=320.75 K-284.65 K=36.1 K$

$Q_1=m\times c\times \Delta T$

$=75.0 g\times 0.91 J/g K\times 36.1 K=2,463.82 J$

2) Heat required to raise the change the state of of fluorocarbon compound from liquid to gas= $Q_2$

$C_2Cl_3F_3(l)\rightarrow C_2Cl_3F_3(g)$

Mass of fluorocarbon compound = m = 75.0 g

Moles of fluorocarbon compound = $\frac{75.0 g}{187.5 g/mol}=0.4 mol$

The heat of vaporization for the compound = $\Delta H_{vap}=27.49 kJ/mol$

$Q_2= \Delta H_{vap}\times 0.4 mol=27.49 kJ/mol\times 0.4 mol=10.996 kJ$

$Q_2=10.996 kJ = 10,996 J$ (1 kJ= 1000 J)

3) Heat required to raise the temperature of fluorocarbon compound from 47.6 °C to 80.30°C= $Q_3$

$C_2Cl_3F_3(g,320.75 K)\rightarrow C_2Cl_3F_3(g,353.45 K)$

Mass of fluorocarbon compound = m = 75.0 g

Specific heat of fluorocarbon compound ,c'= 0.67 J/gK

Change in temperature = $\Delta T'=353.45 K-320.75 K=32.7 K$

$Q_3=m\times c'\times \Delta T'$

$=75.0 g\times 0.67 J/g K\times 32.7 K=1,643.18 J$

The heat required to convert 75.0 g of $C_2Cl_3F_3$ from a liquid at 11.50 °C to a gas at 80.30 °C:

$Q_1+Q_2+Q_3=2,463.82 J+10,996 J+1,643.18 J=15,103 J= 15.103 kJ$