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4.4: Calculations of Enthalpies of Reaction at T ≠ 298 K

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    414049
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    Standard enthalpies of formation are usually reported at room temperature (\(T\) = 298 K), but enthalpies of formation at any temperature \(T'\) can be calculated from the values at 298 K using eqs. (2.4) and (3.13):

    \[ \begin{aligned} dH = C_P dT \rightarrow & \int_{H_{T=298}^{-\kern-6pt{\ominus}\kern-6pt-}}^{H_{T'}} dH = \int_{T=298}^{T'} C_P dT \\ & H_{T'}^{-\kern-6pt{\ominus}\kern-6pt-}- H_{T=298}^{-\kern-6pt{\ominus}\kern-6pt-}= \int_{T=298}^{T'} C_P dT \\ & H_{T'}^{-\kern-6pt{\ominus}\kern-6pt-}= H_{T=298}^{-\kern-6pt{\ominus}\kern-6pt-}+ \int_{T=298}^{T'} C_P dT, \end{aligned} \tag{4.9} \label{4.9} \]

    which, in conjunction with Hess’s Law (Equation \ref{4.8}), results in:

    \[ \Delta_{\text{rxn}} H_{T'}^{-\kern-6pt{\ominus}\kern-6pt-}= \Delta_{\text{rxn}} H_{T=298}^{-\kern-6pt{\ominus}\kern-6pt-}+ \int_{T=298}^{T'} \Delta C_P dT, \tag{4.10} \label{4.10} \]

    with \(\Delta C_P = \sum_i \nu_i C_{P,i}\).

    Exercise \(\PageIndex{2}\)

    Calculate \(\Delta_{\text{rxn}}H\) of the following reaction at 398 K, knowing that \(\Delta_{\text{rxn}}H^{-\kern-6pt{\ominus}\kern-6pt-}\) at 298 K is -283.0 kJ/mol, and the following \(C_P\) values: \(\mathrm{CO}_{(g)}\) = 29 J/(mol K), \(\mathrm{O}_{2(g)}\) = 30 J/(mol K), \(\mathrm{CO}_{2(g)}\) = 38 J/(mol K):

    \[ \mathrm{CO}_{(g)}+\frac{1}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)}, \nonumber \]

    Answer

    Using Equation \ref{4.10} we obtain:

    \[ \Delta_{\text{rxn}} H^{398} = \overbrace{-283.0}^{\Delta_{\text{rxn}}H^{-\kern-6pt{\ominus}\kern-6pt-}} + \int_{298}^{398} ( \overbrace{38}^{C_P^{\mathrm{CO}_2}} -\overbrace{29}^{C_P^{\mathrm{CO}}} -\frac{1}{2}\overbrace{30}^{C_P^{\mathrm{O}_2}} ) \times 10^{-3} dT, \nonumber \]

    which, assuming that the heat capacities does not depend on the temperature, becomes:

    \[ \begin{aligned} \Delta_{\text{rxn}} H^{398} &= -283.0 + \left(38-29-\frac{1}{2}30 \right) \times 10^{-3} (398-298) \\ &= -283.6 \; \text{kJ/mol}. \end{aligned} \nonumber \]

    As we notice from this result, a difference in temperature of 100 K translates into a change in \(\Delta_{\text{rxn}}H^{-\kern-6pt{\ominus}\kern-6pt-}\) of this reaction of only 0.6 kJ/mol. This is a trend that is often observed, and values of \(\Delta_{\text{rxn}}H\) are very weakly dependent on changes in temperature for most chemical reactions. This numerical result can also be compared with the amount that is experimentally measured for \(\Delta_{\text{rxn}}H^{398}\) of this reaction, which is –283.67 kJ/mol. This comparison strongly supports the assumption that we used to solve the integral in Equation \ref{4.10}, confirming that the heat capacities are mostly independent of temperature.


    This page titled 4.4: Calculations of Enthalpies of Reaction at T ≠ 298 K is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Roberto Peverati via source content that was edited to the style and standards of the LibreTexts platform.

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