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.


[![enter image description here](https://i.stack.imgur.com/wETpv.png)](https://i.stack.imgur.com/wETpv.png)


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


[![enter image description here](https://i.stack.imgur.com/Jnrzx.png)](https://i.stack.imgur.com/Jnrzx.png)




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.


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