Question

The compound CF₃CHCl₂ (with a C-Cbond) has been proposed as a substitute for CCl₃ and CCl₂F₂because it decomposes more quickly in the atmosphere and is much Iess liable to reduce the concentration of ozone in the stratosphere. Use the atomization enthalpies and average bond enthalpies from the given data to estimate the standard enthalpy of formation ($\Delta H_i$) of CF₃CHCl₂in the gas phase.

Short answer

The answer is $\Delta H=-608.2kJ$.

Step-by-step solution

Given data

Average Bond enthalpy:

$$\begin{gathered} \Delta H_f(Cl,g)=121.7kJ \\ \Delta H_f(C,g)=716.7kJ \\ \Delta H_f(F,g)=79.0kJ \\ \Delta H_f(H,g)=218kJ \end{gathered}$$

Also given that:

$$\begin{gathered} \Delta H_f(C-Cl)=328kJ \\ \Delta H_f(C-F)=441kJ \\ \Delta H_f(C-H)=413kJ \\ \Delta H_f(C-C)=348kJ \end{gathered}$$

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,H and F.

The reaction would be: $C\left(s\right)+3/2F_2\left(g\right)+1/2H_2\left(g\right)+C{l}_2\to C{F}_{3}CHC{l}_2\left(g\right)$

This reaction is separated into 2 hypothetical processes.

1. All the species on the left are atomized.

2.The atoms combine to make CF₃CHCl₂.

 Step 3: Calculate the first enthalpy change

1) All the species on the left are atomized

The reaction, $C\left(s\right)+3/2F_2\left(g\right)+1/2H_2\left(g\right)+C{l}_2\to C{F}_{3}CHC{l}_2\left(g\right).$.

The enthalpy change for this reaction:$\Delta H_1=2\Delta H_f\left(C\right)+3\Delta H_f\left(F\right)+2\Delta H_f\left(Cl\right)+\Delta H_f\left(H\right)$

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

The given data:

$$\begin{gathered} \Delta H_f(Cl,g)=121.7kJ \\ \Delta H_f(C,g)=716.7kJ \\ \Delta H_f(F,g)=79.07kJ \\ \Delta H_f(H,g)=218kJ \end{gathered}$$

By the data, calculate the $\Delta H_1$.

$$\begin{gathered} \Delta H_1=2\times 716.7kJ+3\times 79.0kJ+2\times 121.7kJ+218kJ \\ =2138.8kJ \end{gathered}$$

Calculate the second enthalpy change

2) The atoms combine to make CF₃CHCl₂.

The reaction, $C\left(s\right)+3F\left(g\right)+H\left(g\right)+2Cl\to C{F}_{3}CHC{l}_2\left(g\right).$

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

This process contains the formation of one C-C, 3 C-F, 2C-Cl, and one C-H bond per molecule.

The given data:

$$\begin{gathered} \Delta H_f(C-Cl)=328kJ \\ \Delta H_f(C-F)=441kJ \\ \Delta H_f(C-H)=413kJ \\ \Delta H_f(C-C)=348kJ \end{gathered}$$

By the data, calculate the $\Delta H_2$.

$$\begin{gathered} \Delta H_2=-(348+3x441+2x328+413) \\ =-2740kJ \end{gathered}$$

Calculate the standard enthalpy of formation

Add the change in enthalpies of the both processes.

$\begin{array}{l}\Delta H=H_1+H_2\\ =2131.8-2740\\ =-608.2kJ\end{array}$