What do the units of enthalpy kJ/mol represent?

Question

Enthalpy is making me slightly confused.

Here $\Delta H$ refers to the change in enthalpy. Does this refer to the enthalpy change of reaction? Are they the same? It seems that change in enthalpy, standard enthalpy change, and standard enthalpy change of reaction are all interchangeable. However, standard enthalpy changes of combustion, neutralisation, formation, etc.. have different definitions. Is this correct?

Now the diagram above shows the reaction between $\ce{Cl2 + H2 -> 2HCl}$. With my understanding, $\Delta H$ refers to the change in enthalpy when one mole of $\ce{Cl2}$ reacts with one mole of $\ce{H2}$ to form two moles of $\ce{HCl}$.

$$\ce{Cl2 + H2 -> 2HCl}\quad\Delta H=x \ \mathrm{kJ/mol}$$

$x$ has some numerical value. I think $\Delta H$ is measured in $\mathrm{kJ/mol}$ as the exact quantities of each element are unknown. However, If we were to have exactly one mole of $\ce{Cl2}$ and one mole of $\ce{H2}$ then the change in enthalpy would be $x\ \mathrm{kJ}$. Without the $\mathrm{/mol}$. So enthalpy can be measured in either $\mathrm{kJ}$ or $\mathrm{kJ/mol}$ right?

Finally what if we have the reaction: $\ce{CH4 + 2O2 -> CO2 + 2H2O}$ where $\Delta H=-890\ \mathrm{kJ/mol}$. What does the per mole now refer to? Because now the reaction is between one mole of $\ce{CH4}$ and two moles of $\ce{O2}$ it doesn't make sense to me to talk about enthalpy change per mole anymore. Could someone explain intuitively what the per mole now represents?

Answer 1

Imagine the reaction:

$$\ce{CH4 +2O2 -> CO2 + 2H2O}$$

Now one can interpret that one molecule of $\ce{CH4}$ reacts with 2 molecules of $\ce{O2}$ to produce one molecule of $\ce{CO2}$ and two molecules of $\ce{H2O}$. It will be equally valid if another person says that,

$$\ce{1 mol CH4 +2 mol O2 -> 1 mol CO2 + 2 mol H2O}$$,

say we can measure the heat released during the reaction, and it can be expressed as a number with a unit Joule. Say this reaction releases 890,000 Joules. Since the heat is released, let us insert a negative sign, i.e., - 890 Joules. Now we can write

$\ce{1 mol CH4 +2 mol O2 -> 1 mol CO2 + 2 mol H2O}$ with -890 kJ.

Can I divide the whole statement by mol? What do we get, now?

$\ce{1 CH4 +2 O2 -> 1 CO2 + 2 H2O}$ with -890 kJ/mol.

So, the per mol in the enthalpy is per mol of the entire reaction. It confirms that the original equation's coefficients were understood to be mol rather than the number of molecules.

Answer 2

Someone explained this in a way that made more sense to me. Consider that each reaction is talking about molecules rather than moles. So now you have 1 molecule of CH$_4$, 2 molecules of O$_2$, and so forth. If you run that reaction 1 mole of times, you get $\Delta$H. If you run it 2 moles of times, you get 2$\Delta$H.

That's the only way it makes sense to me because you don't run the experiment 6X10$^{23}$ times to get $\Delta$H. You only run it once. But you don't run it with 2 molecules of O$_2$. You run it with 2 moles of O$_2$. So that's as though you're running it 6X10$^{23}$ times in a giant batch with 2 molecules of O$_2$ per reaction.