# Values for heat of reaction example

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).

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.

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.

• 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. Mar 8 '20 at 11:02
• @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. Mar 8 '20 at 11:21
• @ 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 $\Delta H$ ? Mar 8 '20 at 14:19
• 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. Mar 8 '20 at 14:31
• 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 Mar 9 '20 at 1:14

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.