kJ per mole or kJ per 2 moles?

Question

From Chemguide:

Here is a simple reaction between hydrogen and oxygen to make water:

$$\ce{2 H2(g) + O2(g) -> 2 H2O(l)} \quad \Delta_\mathrm{r}H^\circ = \pu{572 kJ mol^-1}$$

• First, notice that the symbol for a standard enthalpy change of reaction is $\Delta_\mathrm{r}H^\circ.$ For enthalpy changes of reaction, the $“\mathrm{r}”$ (for reaction) is often missed off — it is just assumed.
• The $“\pu{kJ mol^-1}”$ (kilojoules per mole) doesn't refer to any particular substance in the equation. Instead it refers to the quantities of all the substances given in the equation. In this case, $\pu{572 kJ}$ of heat is evolved when 2 moles of hydrogen gas react with 1 mole of oxygen gas to form 2 moles of liquid water.

Is the measurement unit $\pu{kJ mol^-1}$ correct here? Shouldn't it be measured simply in $\pu{kJ},$ because the $-572$ figure is for the whole reaction, i.e. for 2 moles of the product?

Answer 1

the whole reaction, i.e. for 2 moles of the product?

This is your mistake.

The balanced equation $\ce{2H2 + O2 -> 2H2O}$ does not mean that 2 moles of the product are being formed.

The numbers, called stoichiometric coefficients, only tell you about the ratio in which the reactants react and form products.

That is to say, if I had a system in which I reacted 1 mol of $\ce{H2}$ with 0.5 mol of $\ce{O2}$ to form 1 mol of $\ce{H2O}$, the above equation would still be correct. This is because the ratio of $\ce{H2}:\ce{O2}:\ce{H2O}$ is still $2:1:2$. As long as the ratio is $2:1:2$, the equation will still be correct and applicable.

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What $\Delta_\mathrm{r}H^\circ$ tells you is: how much the enthalpy change would be, if the number of moles reacting were the same as the stoichiometric coefficients. Therefore for the above reaction, saying that $\Delta_\mathrm{r}H^\circ = -572~\mathrm{kJ~mol^{-1}}$ means that

If 2 mol of hydrogen were to react with 1 mol of oxygen to form 2 mol of water, the enthalpy change would be -572 kJ.

That means that if you multiplied the equation by 2, $\Delta_\mathrm{r}H^\circ$ would also be multiplied by 2.

$$\ce{4H2 + 2O2 -> 4H2O} \qquad \Delta_\mathrm{r}H^\circ = -1144~\mathrm{kJ~mol^{-1}}$$

This line tells you: if 4 mol of hydrogen were to react with 2 mol of oxygen to form 4 mol of water, the enthalpy change would be -1174 kJ. That's obviously true because you're reacting twice as much reactants to get twice as many products, so the enthalpy change must be twice as much.

So what is the per mol for

As chemguide said, it is not per mol of anything. The per mol arises from the definition of the reaction enthalpy change, $\Delta_\mathrm{r}H^\circ$. I have another answer describing it in the context of Gibbs free energy here.

However, one way to look at it is to remember that the reaction does not stipulate the actual amounts of reactants and products. That is to say, the value $\Delta_\mathrm{r}H^\circ$ on its own cannot help you find an enthalpy change, unless you tell it exactly how much of the reactants and products you have. If you don't specify whether you start with 2 mol or 1 mol of water, you cannot find $\Delta H$ for the process.

That means you need to multiply $\Delta_\mathrm{r}H^\circ$ by an amount of substance (units mol) to get $\Delta H$ (units kJ). So, $\Delta_\mathrm{r}H^\circ$ must have units of kJ/mol.

What is the actual amount of substance that you are multiplying by then? It is simply the amount of substance of any species reacting, divided by its stoichiometric coefficient in the equation.

Therefore, if you have 6 moles of oxygen reacting, the "amount of substance" that you multiply by would be $6~\mathrm{mol}/1$. The enthalpy change would be

$$\Delta H = (-572~\mathrm{kJ~mol^{-1}})\left(\frac{6~\mathrm{mol}}{1}\right) = -3432~\mathrm{kJ}$$

You have to divide by the stoichiometric coefficient. This is necessary for consistency. For example, if you were only given this $\Delta_\mathrm{r}H^\circ$:

$$\ce{4H2 + 2O2 -> 4H2O} \qquad \Delta_\mathrm{r}H^\circ = -1144~\mathrm{kJ~mol^{-1}}$$

then the "amount of substance" you would multiply by is $6~\mathrm{mol}/2$. You would have

$$\Delta H = (-1144~\mathrm{kJ~mol^{-1}})\left(\frac{6~\mathrm{mol}}{2}\right) = -3432~\mathrm{kJ}$$

Note how the answer comes out to be the same.

But it doesn't have to be for oxygen either. That's what Chemguide means when it says it's not for any particular species in the reaction, because it's applicable to all of them. For example, if I told you that 44 mol of water was formed, then you would use

$$\Delta H = (-1144~\mathrm{kJ~mol^{-1}})\left(\frac{44~\mathrm{mol}}{4}\right) = -12584~\mathrm{kJ}$$

Answer 2

orthocresol's answer covers a lot of good ground, but I want to add another concept that may provide some clarity. In my undergraduate chemical engineering kinetics course, we had to apply calculus to the progression of chemical reactions, so as to understand how fast things were being converted, how much heat was being generated/absorbed, etc. We almost always used a quantity called the extent of reaction, usually represented as $\xi_i$, to track the progress of each reaction i that was occurring.

The extent of a given reaction has units of moles, and is defined for a given reaction as written. So, take the reaction as written in your question:

$$ \ce{2H2(g) + O2(g) \stackrel{\xi_1}{\longrightarrow}2H2O(\ell)}\tag{1} $$

For each mole of reaction extent $\xi_1$, two moles of $\ce{H2}$ and one mole of $\ce{O2}$ are consumed, while two moles of $\ce{H2O}$ are formed.

Now, take the same reaction but with all coefficients multiplied by two, as orthocresol did:

$$ \ce{4H2(g) + 2O2(g) \stackrel{\xi_2}{\longrightarrow}4H2O(\ell)}\tag{2} $$

Here, for each mole of reaction extent $\xi_2$, four moles of $\ce{H2}$ and two moles of $\ce{O2}$ are consumed, while four moles of $\ce{H2O}$ are formed.

So: This all relates to heats of reaction (and entropies of reaction, and free energies of reaction, and ...) in that, really, they are always defined relative to the extent of a specific reaction, with specific and defined stoichiometric coefficients. Further, if one changes the way the reaction is written by multiplying all of the stoichiometric coefficients by a constant factor, the numerical value of the heat of reaction associated with the reaction also is multiplied by that same factor.

Thus, for Reaction $(1)$ I would write:

$$ \Delta H^\circ_{r,1} = -572~\mathrm{kJ\over mol~\xi_1} $$

Similarly, for Reaction $(2)$ I would write:

$$ \Delta H^\circ_{r,2} = -1144~\mathrm{kJ\over mol~\xi_2} $$

But, in both cases, the energy released, say, per mole of $\ce{H2}$ combusted is the same:

$$ -572~\mathrm{kJ\over mol~\xi_1}\times\mathrm{mol~\xi_1\over 2~mol~\ce{H2}} = -286~\mathrm{kJ\over mol~\ce{H2}} \\ -1144~\mathrm{kJ\over mol~\xi_2}\times\mathrm{mol~\xi_2\over 4~mol~\ce{H2}} = -286~\mathrm{kJ\over mol~\ce{H2}} $$

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NOTE: Proper definition of the basis for things like heats of reaction is actually something that is done rather poorly in a lot of cases. I've spent a lot of time at different points in my scientific career trying to sleuth out whether a value is given per mole of $\ce A$, per mole of reaction extent, or whatever.

Answer 3

From Wikipedia of Standard Enthalpy of Reaction:

The standard enthalpy of reaction (denoted $\Delta H_\mathrm{r}^\Theta$) is the enthalpy change that occurs in a system when one mole of matter is transformed by a chemical reaction under standard conditions.

For the above reaction, the stoichiometry of one mole is taken into consideration when one mole of hydrogen and half a mole of oxygen react to form one mole of Water.

For a generic chemical reaction

$\ce{$−v$_{A} A + $−v$_{B} B +\ \dots -> $v$_{P} P + $v$_{Q} Q\ \dots}$

the standard enthalpy of reaction $\Delta H_\mathrm{r}^\Theta$ is related to the standard enthalpy of formation $\Delta H_\mathrm{f}^0$ of the reactants and products by the following equation:

$$\Delta H_\mathrm{r}^\Theta = \sum_B v_B \Delta H_\mathrm{f}^\Theta \left( B \right)$$

You can read more from here — Standard Enthalpy of Reaction