I am currently studying the textbook Mass Spectrometry, third edition, by Jürgen H. Gross. Chapter 2.4.3 Bond Dissociation Energies and Heats of Formation says the following:

Energetics of $\ce{H}^\bullet$ loss from $\ce{CH_4^{+ \bullet}}$ The minimum energy needed to form a $\ce{CH_3^+}$ ion and a hydrogen radical from the methane molecular ion can be estimated from the heat of reaction, $\ce{\Delta H_r}$, of this process. According to Fig 2.6, $\ce{\Delta H_r} = \ce{AE_{(CH_3^+)}} - \ce{IE_{(CH_4)}}$. In order to calculate the missing $\ce{AE_{(CH_3^+)}}$ we use the tabulated values of $\ce{\Delta H_{f(H^\bullet)}} = 218.0 \ \text{kJ mol}^{-1}$, $\ce{\Delta H_{f(CH_3^+)}} = 1093 \ \text{kJ mol}^{-1}$, $\ce{\Delta H_{f(CH_4)}} = -74.9 \ \text{kJ mol}^{-1}$, and $\ce{IE_{(CH_4)}} = 12.6 \ \text{eV} = 1216 \ \text{kJ mol}^{-1}$. First, the heat of formation of the methane molecular ion is determined based on the experimental value of $\ce{IE_{(CH_4)}}$:

$$\ce{\Delta H}_{f(CH_4^{+\bullet})} = \ce{\Delta H_{f(CH_4)}} + \ce{IE_{(CH_4)}} \tag{2.14}$$

$$\ce{\Delta H}_{f(CH_4^{+\bullet})} = -74.9 \ \text{kJ mol}^{-1} + 1216 \ \text{kJ mol}^{-1} = 1141.1 \ \text{kJ mol}^{-1}$$

Then, the heat of formation of the products is calculated from:

$$\ce{\Delta H_{f(prod)}} = \ce{\Delta H_{f(CH_3^+)}} + \ce{\Delta H_{f(H^\bullet)}} \tag{2.15}$$

$$\ce{\Delta H_{f(prod)}} = 1093 \ \text{kJ mol}^{-1} + 218 \ \text{kJ mol}^{-1} = 1311 \ \text{kJ mol}^{-1}$$

Now, the heat of reaction is obtained from the difference

$$\ce{\Delta H_r} = \ce{\Delta H_{f(prod)}} - \ce{\Delta H}_{f(CH_4^{+\bullet})} \tag{2.16}$$

$$\ce{\Delta H_r} = 1311 \ \text{kJ mol}^{-1} - 1141.1 \ \text{kJ mol}^{-1} = 169.9 \ \text{kJ mol}^{-1}$$

The value of $169.9 \ \text{kJ mol}^{-1}$ ($1.75 \ \text{eV}$) corresponds to $\ce{AE_{(CH_3^+)}} = 14.35 \ \text{eV}$, which is in good agreement with published values of about $14.3 \ \text{eV}$ (Fig. 2.7).

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Note: The amount of energy needed to be transferred to the neutral $\ce{M}$ to allow for the detection of the fragment ion $m_1^+$ is called the appearance energy ($AE$) of that fragment ion.

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Taking the values from Fig. 2.7 and using this calculator, we get that $1311 \ \text{kJ mol}^{-1} = 13.588 \ \text{eV}$ and $1386 \ \text{kJ mol}^{-1} = 14.365 \ \text{eV}$.

Unless I have made a mistake or am misunderstanding something, these values do not agree with what the author has presented. It seems that, if we take the values from Fig 2.7, the value of $169.9 \ \text{kJ mol}^{-1}$ ($1.75 \ \text{eV}$) corresponds to $\ce{AE_{CH_3^+}} = 13.588 \ \text{eV}$, which implies that the published value should be around $14.365 \ \text{eV}$.

So has the author made a small mistake here? Or am I misunderstanding something?

I would greatly appreciate it if people would please take the time to review this.

  • $\begingroup$ There is a huge amount of data in this message. But it would be interesting to know exactly what the question is. The title is "Value for heat of reaction". Well ! Which reaction ? There are plenty in the text. $\endgroup$
    – Maurice
    Mar 8, 2020 at 11:02
  • $\begingroup$ @Maurice I am checking (at the end of the post) the author's work here: "The value of $169.9 \ \text{kJ mol}^{-1}$ ($1.75 \ \text{eV}$) corresponds to $\ce{AE_{CH_3^+}} = 14.35 \ \text{eV}$, which is in good agreement with published values of about $14.3 \ \text{eV}$ (Fig. 2.7)." As you can see, the point is that the values that I get are different from those of the author. $\endgroup$ Mar 8, 2020 at 11:21
  • $\begingroup$ @ The Pointer. This is exactly my demand. What is the meaning of this $169.9 kJ mol^{-1}$ ? It corresponds to which $\Delta H$ ? Same question for $1311 kJ mol^{-1}$. This $1311 kJ mol^{-1}$ correspond to which $\Delta H $ ? $\endgroup$
    – Maurice
    Mar 8, 2020 at 14:19
  • $\begingroup$ To answer your first question, it corresponds to $\ce{\Delta H_r} = 1311 \ \text{kJ mol}^{-1} - 1141.1 \ \text{kJ mol}^{-1} = 169.9 \ \text{kJ mol}^{-1}$, which I presumed is written as $\Delta H_r = 170$ in Fig. 2.7. To answer your second question, it corresponds to $\ce{\Delta H_{f(prod)}} = 1093 \ \text{kJ mol}^{-1} + 218 \ \text{kJ mol}^{-1} = 1311 \ \text{kJ mol}^{-1}$. Reading this again, I admit that I am even more confused now about what I did. Do you think I'm interpreting this incorrectly? I'm curious how you interpret it? Sorry for the confusion. $\endgroup$ Mar 8, 2020 at 14:31
  • $\begingroup$ It looks like you’re overlooking that the methane is at -75, so you need to add 75 to 1311 to get the total change in energy that corresponds to AE $\endgroup$
    – Andrew
    Mar 9, 2020 at 1:14

1 Answer 1


The exercise in question is a prediction of the appearance energy of the methyl cation fragment $\ce{CH3+}$. The authors predict that value by estimating the enthalpy for the reaction of methane ionization followed by loss of a hydrogen atom. The enthalpy change estimation is based on published values for heats of formation and bond dissociation energies that have been determined experimentally.

The values presented, however, are rounded to the nearest integer or tenth, and the authors continue to round values during the explanation. For example, 12.6 eV is converted to the closest integer value of 1216 kJ/mol. All values in Fig 2.7 are presented as integers.

There seem to be two possible explanations for the final value being given as 14.35 rather than your calculated value of 14.365. Most likely, the authors have chosen to round to the nearest multiple of 0.05 rather than the nearest tenth or hundredth. Alternatively, they could have used unrounded values throughout the calculation rather than the rounded numbers presented in the step-by-step description. Either way, the difference is insignificant relative to the (im)precision with which the value can be predicted or experimentally determined.

  • $\begingroup$ Thanks for the answer. So there doesn't seem to be any errors in how the author did the calculation? That was what I was mostly worried about, since I don't want to learn something that is actually incorrect. $\endgroup$ Mar 10, 2020 at 2:03

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