# Temperature dependence of reaction enthalpy

Estimate $\Delta_\mathrm{c}H^\circ(500\mathrm{~K})$ for methane by using the data given on the temperature dependence of heat capacities.

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

# My working

First, heat capacities as functions of temperature.

\begin{align} C_{p,m}(\ce{CO2}) &= 44.22 + 8.79\times10^{-3} T - \frac{8.62\times10^{5}}{T^2}\\ C_{p,m}(\ce{H2O}) &= 75.29 + 0 + 0 \\ C_{p,m}(\ce{O2}) &= 29.96 + 4.18\times10^{-3}T - \frac{1.67\times10^{5}}{T^2} \\ C_{p,m}(\ce{CH4}) &= 35.31 \end{align}

Note: a function of temperature dependence wasn't given in the data table for methane, this value is at $298 \mathrm{~K}$. I used this because I did not know what else to do.

Then, I used Kirchhoff's Law:

$$\Delta H(500\mathrm{~K}) = \Delta H(298 \mathrm{~K}) + \int\limits_{298\mathrm{~K}}^{500\mathrm{~K}}\Delta C_{p,m}\,\mathrm{d}T$$

I computed $$\Delta C_{p,m} = \sum_{\mathrm{products}}\nu C_{p,m}-\sum_{\mathrm{reactants}}\nu C_{p,m}$$ (here, I used the molar heat capacities values listed above, adding the functions after multiplying by the appropriate stoichiometric coefficients) and obtained:

$$\Delta C_{p,m} = 16.15 + 0.43\times10^{-3}T - \frac{5.28\times10^{5}}{T^2}$$

Therefore:

\begin{align} &&\int\limits_{298\mathrm{~K}}^{500\mathrm{~K}}\Delta C_{p,m}\,\mathrm{d}T &= 2.581 \mathrm{~kJ~mol^{-1}} \\ &&\Delta H^\circ(298\mathrm{~K}) &= -890 \mathrm{~kJ~mol^{-1}} \\ &\Rightarrow & \Delta H^\circ(500\mathrm{~K}) &= -887.4 \mathrm{~kJ~mol^{-1}} \end{align}

However, my result is incorrect. The correct result is $-803.07\mathrm{~kJ~mol^{-1}}$.

As a reference, the textbook is Physical Chemistry 10th ed by Atkins.

• Note to others reading: this is Exercise 2C.9(a) on page 105 of the text. Oct 25, 2015 at 10:56
• I fixed some $\LaTeX$ issues. There's a couple of things I want to nitpick on. First is the lack of units in your expressions for heat capacity - I know it's all in $\mathrm{J~K^{-1}~mol^{-1}}$ but yeah. Second thing is that you shouldn't include the differential $\mathrm{d}T$ in the equation when you're finding $\Delta C_{p,m}$. The differential changes the meaning of the equation Oct 25, 2015 at 10:58
• Now on to the important part which is the question itself! I redid the whole question and got a value of -799.76 kJ/mol. Oh well... There are a couple of things I would question. The first is your ΔH at 298 K. I used values of $\Delta H_f$ from the appendix and got a standard reaction enthalpy of -802.34 kJ/mol (remember to use H2O(g) and not H2O(l)). Oct 25, 2015 at 11:20
• The next is your calculation of $\Delta C_{p,m}$. 1) For H2O you should use the heat capacity of steam, not water; Table 2B.1 only gives that of water (it says liquids). So I would take the value from Table 2C.5 instead. Next, for the values that you do take from Table 2B.1, don't forget the $10^{-3}$ coefficient in $b$ and $10^5$ coefficient in $c$ (look at the header row). Oct 25, 2015 at 11:22
• @orthocresol: Hi! Thanks for the cleaning up the latex. And I have taken note of everything that you mentioned, and fixed the values in the question to reflect that. It was quite stupid of meto ignore the exponents in the coefficients. The enthalpy at 298K comes from 2c.4, the enthalpy of combustion. (am I not allowed to use that?) If I do, then I get the result stated above. If use -802.34 kJ/mol , however, I do indeed get -799.76 kJ/mol Oct 25, 2015 at 12:24

All enthalpies are taken to be at $298\mathrm{~K}$ if not specified.

The enthalpy of combustion that is shown in Table 2C.4, $\Delta_\mathrm{c} H^\circ = -890\mathrm{~kJ~mol^{-1}}$, corresponds to $\Delta H^\circ$ for the reaction:

$$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$$

which is not the reaction you are interested in (since $\ce{H2O}$ is in the liquid state). Therefore, to calculate the enthalpy of the reaction, you have to use $\Delta_\mathrm{f} H^\circ$ values from Tables 2C.4 and 2C.5:

\begin{align} \Delta H^\circ &= 2\Delta_\mathrm{f} H^\circ(\ce{H2O(g)}) + \Delta_\mathrm{f} H^\circ(\ce{CO2(g)}) - \Delta_\mathrm{f} H^\circ(\ce{CH4(g)}) \\ &= -802.34\mathrm{~kJ~mol^{-1}} \end{align}

as I said in the comments.

Now, even though the question said to use the temperature dependence of heat capacities in Table 2C.1 (in the appendix), the solution manual irritatingly takes the values from Problem 2C.7, on page 106. Here, the heat capacity is expressed in the form $C_{p,\mathrm{m}} = \alpha + \beta T + \gamma T^2$:

$$\begin{array}{c|ccc} \mathrm{Molecule} & \alpha / (\mathrm{J~K^{-1}~mol^{-1}}) & \beta / (\mathrm{mJ~K^{-2}~mol^{-1}}) & \gamma / (\mathrm{μJ~K^{-3}~mol^{-1}}) \\ \hline \ce{CH4(g)} & 14.16 & 75.5 & -17.99 \\ \ce{CO2(g)} & 26.86 & 6.97 & -0.82 \\ \ce{O2(g)} & 25.72 & 12.98 & -3.862 \\ \ce{H2O(g)} & 30.36 & 9.61 & 1.184 \end{array}$$

Therefore, we have:

\begin{align} \Delta C_{p,\mathrm{m}} &= 2C_{p,\mathrm{m}}(\ce{H2O(g)}) + C_{p,\mathrm{m}}(\ce{CO2(g)}) - 2C_{p,\mathrm{m}}(\ce{O2(g)}) - C_{p,\mathrm{m}}(\ce{CH4(g)}) \\ &= \left[21.98 - \left(75.27 \times 10^{-3}\right)(T/\mathrm{K}) + \left(27.262 \times 10^{-6}\right)(T/\mathrm{K})^2\right] \mathrm{~J~K^{-1}~mol^{-1}} \end{align}

and integrating from $298\mathrm{~K}$ to $500\mathrm{~K}$ (I got lazy and used my graphing calculator):

$$\int\limits_{298\mathrm{~K}}^{500\mathrm{~K}}\!\!\Delta C_{p,\mathrm{m}}\,\mathrm{d}T = -731 \mathrm{~J~mol^{-1}}$$

and:

\begin{align} \Delta H^\circ (500~\mathrm{K}) &= \Delta H^\circ (298~\mathrm{K}) + \int\limits_{298\mathrm{~K}}^{500\mathrm{~K}}\!\!\Delta C_{p,\mathrm{m}}\,\mathrm{d}T \\ &= (-802.34 - 0.731) \mathrm{~kJ~mol^{-1}} \\ &= -803.07 \mathrm{~kJ~mol^{-1}} \end{align}

Now that's irritating, because if we use whatever data we can find from the appendix, we get the answer $-799.76\mathrm{~kJ~mol^{-1}}$, as mentioned in the comments. However, I suppose there's two things that can be learnt:

1. Do not assume that the combustion reaction gives $\ce{H2O(g)}$, especially if it is given for $T = 298\mathrm{~K}$.
2. Check the units in the heading of the table.

Your procedure itself is perfectly sound, which is a good thing.

• What is ∆r Cp in general khirchoff equation.....is it like the difference of Cp of product and reactant in a reaction Aug 8, 2018 at 14:48
• @Abhinav yes, but you have to weight it by the stoichiometric coefficient. This is explained in (any edition of) Atkins' Physical Chemistry. Aug 8, 2018 at 17:39