Noting that the bond disassociation energy of $H-O$ is 463 kJ/mol, the standard heat of formation of $H_2O$ is -242 kJ/mol, and the heat of formation for the H* radical is 218 kJ/mol, find the standard heat of formation of $OH$* (the Hydroxyl radical) based on the reaction below:
$$
H_2O_{(g)}+hv\to H_{(g)}+OH_{(g)}
$$

Here, I'm receiving 24kJ/mol, which does not seem to match the generally-accepted value.

Question

Noting that the bond disassociation energy of $H-O$ is 463 kJ/mol, the standard heat of formation of $H_2O$ is -242 kJ/mol, and the heat of formation for the H* radical is 218 kJ/mol, find the standard heat of formation of $OH$* (the Hydroxyl radical) based on the reaction below:
$$
H_2O_{(g)}+hv\to H_{(g)}+OH_{(g)}
$$

Here, I'm receiving 24kJ/mol, which does not seem to match the generally-accepted value.

nincompoop · Answer

$$△H = H_{broken} - H_{formed}$$

nincompoop · Answer

make sure that you're using the correct values from the table also.

anonymous · Answer

Yes, but, go through each step, because I can't get the generally-agreed upon result of $\approx 39\text{ kJ/mol}$.

nincompoop · Answer

consider that in your equation that O exists in nature as diatomic

anonymous · Answer

Wait, wait, wait... what. How will that affect it at all?

anonymous · Answer

I think I see what you mean, but, please, continue.

nincompoop · Answer

you will have $$2H_2$$ and 
$$O_2$$

a few elements exists as diatomic (a kind of allotrope)

anonymous · Answer

Yes...? Do recall that these are free-radicals, not standard-state elements.

nincompoop · Answer

let me try, I didn't really fully work on your problem.

anonymous · Answer

E.g. were these common-state elements (e.g. in solution or in a container), we would have, as a reaction, the disassociation of water into two ions $OH^-$ and $H^+$. This is a photonic reaction, hence we have two radicals, $OH^*$ and an $H^*$.

nincompoop · Answer

I realize that… i didn't see the whole thing earlier

nincompoop · Answer

let me see your calculation?

anonymous · Answer

Sure:
$$
\Delta H=-436+218-(-242)=24\text{ kJ/mol}
$$

nincompoop · Answer

I am going to use my computer, using iPad is annoying with the auto-refresh rate

nincompoop · Answer

brb

anonymous · Answer

I gotcha, haven't been able to use my iPad on OS, too unworkable, personally.

nincompoop · Answer

463 not 436…

anonymous · Answer

Oops, thanks, mate. Still, though, I get an even less valid result. Ideas?

anonymous · Answer

(Whatever "less valid" means) (@nincompoop )

nincompoop · Answer

sorry, dude.  I am trying to confirm the information you've given:
So far, I've only been able to check H2O -242 kJ/mol; I don't know where you got H-O (or O-H) 463 kJ/mol and H^+ 218 kJ/mol

abb0t · Answer

@nincompoop has the right idea. The formula he has provided is missing $\sum$, which pretty much means to add together. but it's pretty much all you need. Remember also to specify the moles in your calculationss
$$\Delta H^o_{reaction} = \sum \Delta H^o_f (products)- \sum \Delta H^o_f (reactants)$$

nincompoop · Answer

assuming we are working with 1 mol. I scoured the CRC handbook and couldn't find OH^- gas and H^+ gas.

abb0t · Answer

hydrogen gas is zero.

nincompoop · Answer

I just want to make sure that we start from correct data.

nincompoop · Answer

zero if it were an aqueous and in allotropic H2 form.  He reiterated that it is the radical form H^+

abb0t · Answer

standard enthalpy of formation for hydrogen gas, or for any free element in its standard state for that matter, is by definition zero.

anonymous · Answer

Nah, I have it from a paper, although the values I was given are slightly skewed, but, here, @nincompoop .

As for @abb0t , yes, this does not deal with what you mean (I believe it is something like $1/2H_2$, et al). These are free-radicals, not standard-state elements.

anonymous · Answer

Oh, and, be careful, @nincompoop , these aren't ions, they're radicals (it's not quite $H^+_{(g)}$, it's simply $H_{(g)}$, or, to be more clear, it is written $H^*$).

nincompoop · Answer

then it just simply appears that we are using different data sets, which is the subject of this paper. it is good to know, that even the accepted values may be subject to scrutiny.

nincompoop · Answer

I will fully read the paper after my semester is over.   This is beyond my scope though

abb0t · Answer

I don't see why you're looking for standard heat for formation for a radical. Radicals are intermediate products.

anonymous · Answer

Not always, especially when considering vacuums (containing only the matter given) or other cases that are outside of the Earth's atmosphere (as in the example above).

anonymous · Answer

I can close this.

Note that the value given in the test of $D_0(H-OH)$ is wrong, since:
$$
D_0(H-OH)\ne D_0(OH)
$$Using the correct value of:
$$
D_0(H-OH)=498.7\text{ kJ/mol}
$$Yields the correct result after manipulation, which is:
$$
D_0(H-OH)+\Delta H_f^0(H_2O)-\Delta H_f^0(H)=\Delta H_f^0(HO)
$$