Example 5.13: Stepwise Calculation of $\mathrm{ΔHf°}$ Using Hess’s Law

Determine the enthalpy of formation, $\mathrm{ΔHf°}$, of $\ce{FeCl3(s)}$ from the enthalpy changes of the following two-step process that occurs under standard state conditions:

$\ce{[1]~~Fe(s)}$$\ce{ + }$$\ce{Cl2(g)}$$\ce{->}$$\ce{FeCl2(s)~~~~ΔH° = -341.8kJ/mol}$$\ce{ }$

$\ce{[2]~~FeCl2(s)}$$\ce{ + }$$\ce{1/2Cl2(g)}$$\ce{->}$$\ce{FeCl3(s)~~~~ΔH° = -57.7kJ/mol}$$\ce{ }$

Solution

$ΔH°_{\mathrm{rxn1}}$ $= -341.8\ \frac{\mathrm{kJ}}{\mathrm{mol}}$

$ΔH°_{\mathrm{rxn2}}$ $= -57.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}$

We are trying to find the standard enthalpy of formation of $\ce{FeCl3(s)}$, which is equal to $\mathrm{ΔHf°}$ for the reaction:

$\ce{[3]~~Fe(s)}$$\ce{ + }$$\ce{3/2Cl2(g)}$$\ce{->}$$\ce{FeCl3(s)}$$\ce{ }$

If we add the two reactions for which we know the enthalpy, we get:

$\ce{Fe(s)}$$\ce{ + }$$\ce{Cl2(g)}$$\ce{ + }$$\ce{FeCl2(s)}$$\ce{ + }$$\ce{1/2Cl2(g)}$$\ce{->}$$\ce{FeCl2(s)}$$\ce{ + }$$\ce{FeCl3(s)}$$\ce{ }$

To get the net reaction, remove the $\ce{FeCl2}$ occuring both as reactnat and product, and add up the $\ce{Cl2}$ occuring twice in the list of reactant to see that this is the reaction we need:

$\ce{Fe(s)}$$\ce{ + }$$\ce{3/2Cl2(g)}$$\ce{->}$$\ce{FeCl3(s)}$$\ce{ }$

According to Hess's Law, the reaction enthalpy of this reaction is the sum of the two reactions with known $ΔH$ values:

$ΔH°_{\mathrm{rxn3}}$ $= ΔH°_{\mathrm{rxn1}} + ΔH°_{\mathrm{rxn2}}$

$\ \ \ =-341.8\ \frac{\mathrm{kJ}}{\mathrm{mol}} + -57.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}$

$\ \ \ =-399.5\ \frac{\mathrm{kJ}}{\mathrm{mol}}$

$\ce{Fe(s)}$$\ce{ + }$$\ce{3/2Cl2(g)}$$\ce{->}$$\ce{FeCl3(s)~~~~$ΔH° = -399.5$ kJ/mol}$$\ce{ }$

The enthalpy of formation, $\mathrm{ΔHf°}$, of $\ce{FeCl3(s)}$ is -399.5 kJ/mol.

Example 5.13: Stepwise Calculation of \(\mathrm{ΔHf°}\) Using Hess’s Law

Solution