# How do scientist obtain the value of enthalpy of formation? [closed]

Do they do it experimentally or do/can they use bond enthalpy to figure it out? For example $\ce{C2H4 + H2 -> C2H6}$. The enthalpy of formation, $\Delta_fH$ of formation for the product is listed as -84, however if I use bond enthalpy to calculate the $\Delta_fH$, it gives be a value of -222, which is considerably different. Shouldn't the difference between bonds formed and bonds broken account for the heat given off my the reaction? With this discrepancy I can only conclude that I mixed up the two concept.

Also if $\ce{C2H4}$ enthalpy of formation $\Delta_fH$ equals to 52, $\ce{H2}$ = 0 and $\ce{C2H6}$ = -84, the difference between the product and reactant is -136. I am confused about the 3 numbers (-84, -222 and -136) and their significance.

Also, how is $\Delta_fH^\circ$ different from $\Delta_RH^\circ$ of reaction? When do I use each one?

I would really appreciate it if someone can clear this up concisely

## closed as too broad by Mithoron, Pritt Balagopal, Jon Custer, M.A.R., Jannis AndreskaJul 13 '17 at 16:45

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• You seem to have more than one question here. Try to limit your post to one question that addresses your main area of misunderstanding and, if that still doesn't answer everything, ask a new question. – Tyberius Jul 6 '17 at 20:56
• @Jim Roberts Welcome to ChemSE. For future reference, be invited to familiarize yourself with mathjx and \mhchem that will help you considerably in formatting questions, as well as answers on ChemSE. To start, have a look at this compilation (chemistry.meta.stackexchange.com/questions/86/…); and graduate, at your pace, with this (chemistry.meta.stackexchange.com/questions/3044/…). – Buttonwood Jul 6 '17 at 21:25
• You'll likely get more traction on this question if you provide a table of the relevant values with units. People are less likely to help if they have to look everything up to double check your work. – Zhe Jul 6 '17 at 21:48

There is a consent to start with: the standard enthalpies of formation of elements (like $\ce{H2}$), at standard conditions ($\pu{1 bar}$, $\pu{298.15 K}$) are set to zero. If there are several forms of one element, than the most stable polymorph is chosen as reference (except for phosphor, ref).

More generally, to determine the $\Delta{}_fH^\circ{}$ of other compounds, you may embark experiments by calorimetry. By means of a calorimeter, you are able to record the energetic changes that occur in the course of a reaction. For example for the exhaustive combustion of carbon with oxygen:

$$\ce{C + O2 -> CO2}$$

all the heat recorded is due to the formation of carbon dioxide. Since this reaction equation was balanced that it equals only one mol of product ($\ce{CO2}$), the more general standard reaction enthalpy ($\Delta_{\text{R}}H^\circ{}$) equals to the standard enthalpy of formation ($\Delta_fH^\circ{}$). And by one experiment, after an other, the body of experimentally determined enthalpies grows.

Based on other experiments than combustion, one was able to determine how much energy is necessary to break a bond between atoms which is called bond dissociation energy. Conversely, this value expresses how much energy over all is released if two atoms establish a bond with each other. In some instances, these energies are (comparatively) small, for example single bonded $\ce{Cl-Cl}$ ($\pu{242 kJ/mol}$), in other instances (comparatively) large, like triple bonded $\ce{N#N}$ ($\pu{945 kJ/mol}$).

If one knows the structure of a compound, which atoms are connected with each other and how (single, double, triple bond), than these bond dissociation energies may be used to estimate the standard enthalpy of formation of the new compound, for example by an increment system. Alternatively, by virtually setting up for example a combustion reaction of exactly one mole of the new compound into $x$ moles of $\ce{H2O}$ and $y$ moles of $\ce{CO2}$. Then, the calculation to perform is

$$\Delta_{\text{R}}H^\circ{} = \sum_{\text{all products}}{\Delta_fH^\circ{}} - \sum_{\text{all starting materials}}{\Delta_fH^\circ{}}$$

where $\Delta_{\text{R}}H^\circ{}$ again is the overall reaction enthalpy. Since these estimations are based on some empiric data, they do not a substitute for an experimentally determined value for your newly prepared compound. Often, increment systems leave some of inter- and intramolecular interactions unconsidered (e.g., ring strain, possible existence of polymorphism) which then yields a difference between the estimated, and the actual value of $\Delta_{\text{R}}H^\circ{}$ or $\Delta_fH^\circ{}$.