# How to calculate an enthalpy change for this reaction

I've got an assignment to solve problem number 4 in this worksheet, that is to calculate change in enthalpy for reaction: $$\ce{CHCl3 + O -> COCl2 + HCl}$$

How do I solve it? Why is oxygen $\ce{O}$, and not $\ce{O2}$?

I figured out that, it is not a combustion reaction, as combustion of any hydrocarbon will lead to formation of $\ce{CO2}$ and $\ce{H2O}$, as per this page.

Well, what I have at the beginning is not exactly a hydrocarbon, but rather, a halogenated hydrocarbon. Yet, no water and no carbon dioxide is produced. So as I mentioned above, I consider it is not a combustion. Any thoughts on this? Could it still be a combustion? Why or why not?

When I calculate $\Delta H$ of this reaction, the total outcome depends on using reaction as it is shown in the worksheet, or correcting $\ce{O}$ to $\ce{O2}$ and balancing it. Which option, do you think, is preferable, and why?

• Also, for general reference on $\LaTeX$, check this meta post: meta.math.stackexchange.com/q/5020 – user5764 Feb 27 '16 at 20:24
• Why should be O$_2$ ? Combustion reactions are vaguely defined, but many times refers to reactions with $O_2$, I would not call this a combustion reaction but as definition is not accurate I can give strong arguments. It could be called combustion if it a fast and very exergonic reaction (I don't know this case) but it is no common to call this way reaction with radicals. I can not view the worksheet. – user1420303 Mar 3 '16 at 23:12

To preface, the reason that this is not a combustion reaction is because this is a redox reaction that occurs at room temperature. In fact, chloroform is noncombustible.

Though at first glance, this may simply appear to be an unbalanced equation that could be rewritten: $$\ce{CHCl3 + 1/2O2 -> COCl2 + HCl}$$

It is better to take it for what the question presents it as, an oxygen atom, so I would therefore recommend you to leave the equation as:

$$\ce{CHCl3 + O -> COCl2 + HCl}$$

Though $\ce{O2}$ is a pure element, and its $\Delta H^0_\mathrm f=0\:\mathrm{kJ\:mol^{-1}}$, for the atom $\ce{O}$, $\Delta H^0_\mathrm f=58.99\:\mathrm{kJ\:mol^{-1}}^{[1]}$ (provided thanks to user1420303).

To calculate $\Delta H$, the equation is: $$\Delta H=\Delta H^0_\mathrm f(\ce{COCl2}) + \Delta H^0_\mathrm f(\ce{HCl})-\Delta H^0_\mathrm f(\ce{CHCl3})-\Delta H^0_\mathrm f(\ce{O})$$

Though enthalpy is a state function, this question is asking you to calculate $\Delta H$ from a specific set of reactants, and their individual energies must be considered. There is more than just one way to calculate $\Delta H$, and an analysis of the $\Delta H^0_\mathrm f$ (the enthalpy of formation of a compound from its constituent elements) for each compound accounts for which bonds are being broken and formed, and will yield (about) the same results as a bond by bond analysis. I find this method to be preferable because it bond by bond analyses are based on average bond enthalpies, while $\Delta H^0_\mathrm f$ is often painstakingly measured for each compound, and will yield the most accurate $\Delta H$ calculation.

[1] Curtiss, L. A.; Raghavachari, K.; Redfern, P. C.; Pople, J. A. Assessment Of Gaussian-2 and Density Functional Theories for the Computation of Enthalpies of Formation. The Journal of Chemical Physics J. Chem. Phys. 1997, 106, 1063.

• delta H of formation and bond energy are two different things. One still needs to invest energy to break double bond between two oxygen atoms and this energy is not equal to zero. – Sleepy Hollow Feb 29 '16 at 17:19
• Delta H of reaction = sum of bonds energy for bonds broken - sum of bonds energy for bonds made. There is no delat H of formation in this equation. – Sleepy Hollow Feb 29 '16 at 17:23
• Regardless of what bonds are broken in a specific reaction, enthalpy is a state function. An analysis of the $\Delta H^0_f$ for each compound (which tells you the enthalpy of formation from the constituent elements of each molecule) will yield the same results as calculating which specific bonds are being broken and formed. There is more than just one way to calculate $\Delta H$. – ringo Feb 29 '16 at 17:25
• This answer is not correct in that $O$ is not the same that $(1/2)O_2$ – user1420303 Mar 3 '16 at 23:04
• @user1420303 Are you trying to say half of a diatomic molecule isn't a single atom? If so I don't know what to tell you. Either way, it's only a stoichiometric concept. In the mechanism for this reaction, the $\pi$-bond in diatomic oxygen is homolytically split, and then reacts via a chain radical mechanism. Never in this reaction does there exist a true oxygen atom. – ringo Mar 3 '16 at 23:47

Let it remain as $\ce{O}$. Using $\ce{O}$, the bonds that are formed are:

1. 1 $\ce{C=O}$ bond
2. 1 $\ce{H-Cl}$ bond

The bonds that are broken are,

1. 1 $\ce{C-H}$ bond
2. 1 $\ce{C-Cl}$ bond

Using the above information and the formula given in your assignment,

$$\Delta_\mathrm{R} H=[\text{energy used for breaking bonds}]-[\text{energy used for forming bonds}]$$

you can find the enthalpy change.

Yes. The reaction given is not a combustion reaction. However, it is not necessary that the products are carbon dioxide and water vapour.

One interesting thing that you can note is that, chloroform stored in presence of oxygen will slowly convert to phosgene and this can take place at room temperature.

Anyways, the main intention of the question is simple and only asks you to calculate the enthalpy change.

• I don't think $\ce{C=O} \approx 2(\ce{C-O})$. – SendersReagent Feb 29 '16 at 16:44
• Eh... They're not too far off. – SendersReagent Feb 29 '16 at 16:48
• @DGS my mistake. I thought $\ce{C = O}$ bond energy was not given. So I thought I could make an approximation :) – Aditya Dev Feb 29 '16 at 16:58

How do I solve it? Why is oxygen O, and not O2?

Because $\ce{O2}$ does not react.$\ce{O2}$ gets dissociated into $\ce{2O}$ and then this oxygen atoms reacts.Remember about bond dissociation enthalpy?

Diatomic molecule is one that only contains two atoms. They could be the same (for example, Cl2) or different (for example, HCl). The bond dissociation enthalpy is the energy needed to break one mole of the bond to give separated atoms - everything being in the gas state.

• That is exactly my point. If I use O2, I have to consider energy I have to use for bond breaking in O2, Outcome (in terms of endothermic or exothermic) is different, if I use it with O, and if I use it with O2, and then balance equation. – Sleepy Hollow Feb 29 '16 at 17:25