Question

A second CFC used as a refrigerant and in aerosols is CCl₃F. Use the atomization enthalpies and average bond enthalpies from the given data to estimate the standard enthalpy of formation $\left(\Delta H_f^{\circ }\right)$of this compound in the gas phase.

Short answer

$Theansweris\Delta H=-264kJ.$

Step-by-step solution

Given data

Average Bond enthalpy:

$$\begin{gathered} \Delta H_f(Cl,g)=121.7kJ \\ \Delta H_f(C,g)=716.7kJ \\ \Delta H_f(F,g)=79.0kJ \\ \Delta H_f(C-Cl)=328kJ \\ \Delta H_f(C-F)=441kJ \end{gathered}$$

Concept of Standard enthalpy

The standard enthalpy of formation is the enthalpy change for the process of formation from its elements in the standard conditions.

In this case, those elements are C, Cl and F.

The reaction would be:

$C\left(s\right)+3/2"C{l}_2\left(g\right)+1/2F_2\left(g\right)\to CC{l}_{3}F\left(g\right)$

This reaction is separated into 2 hypothetical processes.

1. All the species on the left are atomized.
2. The atoms combine to make CCl₃F.

Calculate the first enthalpy change

1) All the species on the left are atomized

The reaction, $C\left(s\right)+3/2C{l}_2\left(g\right)+1/2F_2\left(g\right)\to C\left(g\right)+3Cl\left(g\right)+F\left(g\right).$.

The enthalpy change for this reaction: $\Delta H_1=\Delta H_f\left(C\right)+3\Delta H_f\left(Cl\right)+\Delta H_f\left(F\right)$

There are no reactants involved because the standard enthalpy of them is equal to 0.

The given data:

$$\begin{gathered} \Delta H_f(Cl,g)=121.7kJ \\ \Delta H_f(C,g)=716.7kJ \\ \Delta H_f(F,g)=79.0kJ \end{gathered}$$

By the data, calculate $\Delta H_1$.

$$\begin{gathered} \Delta H_1=716.7+3x121.7+79.90 \\ =1161kJ \end{gathered}$$

Calculate the second enthalpy change

2) The atoms combine to make CCl₃F.

The reaction, $C\left(s\right)+3Cl\left(g\right)+2F\left(g\right)\to CC{l}_{3}F\left(g\right)$.

The enthalpy of this step can be estimated using the bond enthalpies.

This process contains formation of 3C-Cl and one C-F bond per molecule.

The given data:

$\begin{array}{l}\Delta H_f(C-Cl)=328\mathrm{kJ}\\ \Delta H_f(C-F)=441\mathrm{kJ}\end{array}$

By the data, calculate role="math" localid="1663679162787" $\Delta H_2$.

$$\begin{gathered} \Delta H_2=-3\times 328kJ-441kJ \\ =-1425kJ \end{gathered}$$

Calculate the standard enthalpy of formation 

Add the change in enthalpies of the both processes.

$\begin{array}{l}\Delta H=\Delta H_1+\Delta H_2\\ =1161-1425\\ =-264kJ\end{array}$