# Calculating Enthalpy of formation versus Calculating Enthalpy of a reaction not occurring at standard conditions

My understanding of Hess's law and its use in determining the change of enthalpy of a reaction has been challenged recently.

Up until recently, I thought that the only way to calculate a chemical reaction's change of enthalpy was to use its intermediate chemical reactions and their associated enthalpies (usually provided in a question). So, I would just re-arrange the intermediate reactions so that when combined the compounds and elements would be located on the appropriate side of the yield sign (matching the main reaction). Additionally, I would alter their enthalpies at the same time. Finally, I would add all of the intermediate reaction enthalpies and their sum would be the correct change in enthalpy of the main reaction.

However, I learned that when calculating the change in enthalpy of formation you do not simply add all of the other intermediate reaction's change in enthalpy. Instead, after arranging the other reactions appropriately and altering their mole coefficients and making the same changes to their enthalpies, you are supposed to add the reactants together and the products together and then subtract the reactants from the products. Why? This seems to ignore the fact that energy and enthalpy are state functions. Why must I combine the enthalpies of formation in this particular way if enthalpy is path independent?

I am only studying chemistry at a college level. Though at the college level, I am only studying the basics of chemistry. Please consider this when replying to my question. This community is very smart and seems to be, generally, at a level of understanding far above my own. So, again, please do your best to explain the answer at my level.

Thank you.

When I attempt to solve this question using the first method I get $$\pu{-1.09E4 kJ}$$. The correct answer is $$\pu{-1.2E4 kJ}$$.

• Both methods will give you the same answer whether you alter the reactions and add or just subtract the formation enthalpy of reactants from the products. You can try for an example. Commented Jan 5, 2022 at 4:58
• I recently tried to apply the first method I described to a Enthalpy of Formation question, it did not work. I want to share a picture of the question but this comment box won't let me paste it. The best I can do is. . . But, an example would be Find the change in enthalpy for the following reaction: S + O2(g) --> SO2(g). Then the following two equations are provided: S + 3/2O2 --> SO3 (DeltaH = -395 kJ); 2SO2 _ O2 --> 2SO3 (Delta = -198.2 kJ). An example of Enthalpy of formation would be: Using a table determine the standard enthalpy of formation of NO2 (4NH3 + 7O2 --> 4NO2 + 6H2O) Commented Jan 5, 2022 at 22:44
• You can actually edit your question to include all this info and also an image. Commented Jan 6, 2022 at 4:41

As already mentioned by me in the comments, both the approaches are exactly the same.

To explain this point, I would like to find the numerical value of the question posted by you in the comments using both methods.

# By Manipulating the Equations:

## Given:

\begin{align} \begin{array}{ c c} \ce{S + 3/2O2 -> SO3 } & ΔH = \pu{-395 kJ} \\ \ce{2SO2 + O2 -> 2SO3} & ΔH = \pu{+198.2 kJ} \\ \end{array} \end{align}

## To find:

\begin{align} \begin{array}{ c c} \ce{S + O2 -> SO2 } & ΔH = \pu{?? kJ} \\ \end{array} \end{align}

## Method:

We can manipulate the given reactions in the following way to obtain the result we want:

\begin{align} \begin{array}{ c c} \ce{S + 3/2O2 -> SO3 } & ΔH = \pu{-395 kJ} \\ \ce{SO3 -> SO2 + 1/2O2} & ΔH = \pu{+\frac{198.2}{2} kJ} \\ \hline \ce{S + O2 -> SO2} & ΔH =\pu{ \frac{198.2}{2} -395 = -295.9 kJ} \\ \end{array} \end{align}

By this method we get, $$ΔH =\pu{-295.9 kJ}$$.

# By using Enthalpy of Formation:

## Given:

\begin{align} \begin{array}{ c c} \ce{S + 3/2O2 -> SO3 } & ΔH = \pu{-395 kJ} \\ \ce{2SO2 + O2 -> 2SO3} & ΔH = \pu{+198.2 kJ} \\ \end{array} \end{align}

## To find:

\begin{align} \begin{array}{ c c} \ce{S + O2 -> SO2 } & ΔH = \pu{?? kJ} \\ \end{array} \end{align}

## Method:

We can see that we need to find the enthalpy of formation of $$\ce{SO2}$$ in this question, given the enthalpy of formation of $$\ce{SO3}$$ and enthalpy of reaction of a reaction including $$\ce{SO2}$$ and $$\ce{SO3}$$.

For, $$\ce{2SO2 + O2 -> 2SO3}$$ we can say:

\begin{align} \Delta H_r &= 2\Delta H_{f}^\ce{SO3} - 2\Delta H_{f}^\ce{SO2} - \Delta H_{f}^\ce{O2}\\ 2\Delta H_{f}^\ce{SO2} &= 2\Delta H_{f}^\ce{SO3}- \Delta H_{f}^\ce{O2} - \Delta H_r\\ \Delta H_{f}^\ce{SO2} &= \Delta H_{f}^\ce{SO3}- \frac{\Delta H_{f}^\ce{O2}}{2} - \frac{\Delta H_r}{2}\\ \Delta H_{f}^\ce{SO2} &= \pu{(-395 + \frac{198.2}{2} - 0) kJ}\\ \Delta H_{f}^\ce{SO2} &= \pu{-295.9 kJ}\\ \end{align}

By this method also we get, $$ΔH =\pu{-295.9 kJ}$$.

This shows how both the methods are same. The former is a more visual way of thinking about the problem while the latter is a more mathematical approach.

# Additional Calculations after the question was updated

Now you want to find $$\Delta H_r$$ of the following reaction:

$$\ce{2C8H18 + 25O2 -> 16CO2 + 18H2O}$$

Using the $$\Delta H_f$$ of octane, carbon dioxide and water.

You can rearrange the enthalpy of formation reactions in the following way to get the desired outcome:

\begin{align} \begin{array}{ c c} \ce{2C8H18 -> 16C + 18H2 } & ΔH = \pu{(269 \times 2) kJ} \\ \ce{16C + 16O2 -> 16CO2} & ΔH = \pu{-(394 \times 16) kJ} \\ \ce{18H2 + 9O2 -> 18H2O} & ΔH = \pu{-(286 \times 18) kJ} \\ \hline \ce{2C8H18 + 25O2 -> 16CO2 + 18H2O} & ΔH_r ={(269 \times 2) -(394 \times 16) -(286 \times 18)}\pu{ kJ} \\ \end{array} \end{align}

Upon solving we get $$\Delta H_r=\pu{-1.0914E4 kJ}$$. Even if we would have used the second method we would have gotten the same result.

This is the correct answer to the question and not $$\pu{-1.2E4 kJ}$$.

The solution you posted later in the question is incorrect because it assumes that the enthalpy of formation of octane is positive whereas it is provided to be negative.

• I edited my initial question above to now include a picture of the exact question I am struggling with. Can you give the question a try? Thanks for your help so far. Commented Jan 6, 2022 at 11:45
• If you find the answer helpful you can upvote it and accept the answer. 😊 Commented Jan 6, 2022 at 12:13
• Thank you for your sustained effort in explaining the procedure of solving changes in enthalpy. I just upvoted you. Before I accept your answer, can you take a look at the explanation of the answer I just added to my original question? What do you think about it? Commented Jan 6, 2022 at 22:35
• The photo you uploaded is only half the process can you please upload the complete image. Commented Jan 7, 2022 at 2:25
• My bad. The entire explanation is there now. Commented Jan 7, 2022 at 10:52