Temperature dependence of reaction enthalpy
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:

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

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:

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

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 $C_{p,\mathrm{m}} = \alpha + \beta T + \gamma T^2$:

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

Therefore, we have:

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

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:

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

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:


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


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