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So I understand that if twice as much of the reactants are present, then twice as much energy is released. But isn’t the energy released per mole of reactant still the same? You are just scaling up both sides of the equation, I.e. the ratios are still the same, so how is it that your are producing more energy per unit of mass of reactants?

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    $\begingroup$ Chemistry SE prefers plain text titles. For formatting of chem/math formulas/expressions, use can use MathJax with mhchem extension \ce{}. Using photos/screenshots of text instead of typing text itself is discouraged. The image text content cannot be indexed nor searched for, cannot be reused nor referred in answers. $\endgroup$
    – Poutnik
    Jan 20 at 7:26
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    $\begingroup$ These values of $\Delta H$ is given in ... per mole of ..... of what is written in the equation. For your last equation, it would be "per mole of $3$ times what is written". And as it is non-sense to say "per mole of $3$ $\ce{CO2}$", it is better to say "per $3$ moles of $\ce{CO2}$ or “for $ 3$ moles of $\ce{CO2}$" $\endgroup$
    – Maurice
    Jan 20 at 11:12
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By convention, the values given for the changes in thermodynamic state functions $(\Delta G, \Delta H, \Delta S, \Delta A, \Delta U$, etc.) are for the reaction as written. What's confusing is that the "per mole" in the units does not mean per mole of any given reactant or product. Rather, it means per the number of moles* of each reactant and product specified in this reaction, where the number of moles is equal to the compound's stoichiometric coefficient.

Thus the way to interpret the "per mole" for the value of $\Delta H$ for the first reaction is:

This is the enthalpy change per 1 mole of methanol consumed, per 3/2 mole of oxygen consumed, per 1 mole of carbon dioxide produced, and per 2 moles of water produced.

If the reaction were instead written: $$\ce{2CH_3OH(l) +3 O_2(g) -> 2 CO_2(g) + 4 H_2O(l)},$$

then $\Delta H$ would be 2 x –726 kJ/mol = –1452 kJ/mol, and the "per mole" would mean per 2 moles of methanol consumed, per 3 moles of oxygen consumed, per 2 moles of carbon dioxide produced, and per 4 moles of water produced.

I.e., more formally, what kJ/mol means here is, for any of the species in the reaction (whether reactant or product):

$$\frac{ \text{kJ}}{\text{mol} \circ \text{stoichiometric coefficient }}$$

[The stoichiometric coefficient is dimensionless.]

*To those who wish to complain that I am using the phrase "number of moles" instead of "moles", I will argue that here it does make sense. For instance, if I were were talking about liters instead of moles, I would still say the "number of liters" specified for each reactant and product. Or if the stoichiometric coefficients told us how many dozens of atoms and/or molecules were involved, then I would say "number of dozens," etc.

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  • $\begingroup$ Hm, when the reaction enthalpy is calculated, the result is IMHO in kJ, not in kJ/mol. $$\Delta H_\mathrm{r}[\mathrm{kJ}]=\sum {\Delta H_\mathrm{form,prod,i}[\mathrm{kJ/mol}] \cdot n_\mathrm{i}[\mathrm{mol}]} - \sum {\Delta H_\mathrm{form,react,j}[\mathrm{kJ/mol}] \cdot n_\mathrm{j}[\mathrm{mol}]},$$ and as you say, for the reaction as it is written. $\endgroup$
    – Poutnik
    Jan 20 at 7:35
  • $\begingroup$ @Poutnik Yeah, when I started to write my answer I thought the use of per mole was wrong. But then I looked in a few of my pchem texts, and noticed some do adopt this usage. So I tried to understand why, and realized they are trying to communicate, explicitly, that the stoichiometric coefficients refer to moles rather than particles. Thus one could argue either for or against including the per mole. $\endgroup$
    – theorist
    Jan 20 at 8:18
  • $\begingroup$ If used kJ/mol, it should be specified "per mol of ...". In this case it would be methanol, but generally it is unclear. And in such a case, both equation sides would be formally divided by molar amount of the referred substance. $\endgroup$
    – Poutnik
    Jan 20 at 8:35
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    $\begingroup$ @Poutnik That's not what kJ/mol means here. It means $\frac{ kJ}{mol \circ \text{ stoichiometric coefficient }}$. [I'll add this to my answer.] Thus there is no need to specify "per mol of", since it applies to all species—both reactants and products. $\endgroup$
    – theorist
    Jan 20 at 9:17
  • $\begingroup$ Yes, this is the good way. It is analogous to the reaction rates. $\endgroup$
    – Poutnik
    Jan 20 at 10:23

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