Suppose a reaction is occuring at constant temperature. Then $\mathrm dH=nC_{p}\mathrm dT = nC_{p} × 0=0$.

Please explain what the mistake is that I have done. Clearly, there are many reactions which take place at constant temperature and have a definite enthalpy change. But according to the equation $\mathrm dH=nC_{p}\mathrm dT$, if temperature $T$ is constant then $\mathrm dT =0$ and $\mathrm dH = 0$. But this is clearly not true. There are reactions with constant temperature and $\mathrm dH≠0$. Please explain what is wrong?

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    $\begingroup$ dH = nCpdT is true only at constant pressure.. dH = dU - dW $\endgroup$ Commented Apr 16, 2021 at 11:49
  • $\begingroup$ Please refrain from using MathJax in titles, we generally discourage it here. $\endgroup$ Commented Apr 16, 2021 at 12:09
  • $\begingroup$ @SafdarFaisal $\mathrm dH=nC_{p}\mathrm dT$ is valid for all cases, and not just at constant pressure. Of course, this is with the assumption that Cp is calculated for an ideal gas. check here: physics.stackexchange.com/questions/203605/delta-h-c-p-delta-t $\endgroup$
    – C_Lycoris
    Commented Apr 16, 2021 at 12:46
  • $\begingroup$ You are aware that enthalpy is a function not only of temperature and pressure, but also, for a multicomponent system, the amounts of the various components present (which change during a chemical reaction), right? $\endgroup$ Commented Apr 16, 2021 at 14:52

3 Answers 3


The only way that an exothermic reaction can occur with "constant temperature" is if the heat generated is constantly removed in some way. Even if the heat removal is essentially instantaneous, it is incorrect to say that there is no temperature change. Rather there is an infinite number of infinitely small temperature changes as the system heats up by an amount dT and then is returned to the set T. If you were able to measure these infinitely small changes and also to know the instantaneous Cp (which changes as the reaction composition changes), you could sum them up and get a non-zero value for the enthalpy change. Obviously, that's not a practical approach.

Instead, to determine the enthalpy change experimentally, it is easiest to perform the reaction within an insulated apparatus such that the temperature is not constant and the change in temperature can be used to determine the enthalpy change.


For a multicomponent reactive system you would rather write the enthalpy differential as

$$ d(nH) = \left(\frac{\partial (nH)}{\partial T}\right)_{P,n_i}dT + \left(\frac{\partial (nH)}{\partial P}\right)_{T,n_i} dP + \sum_{i = 1}^N \left(\frac{\partial (nH)}{\partial n_i}\right)_{P,T,n_{j\neq i}}dn_i $$

Where $H$ is the molar enthalpy. At constant temperature and pressure, $dP = 0$ and $dT = 0$, therefore

$$ d(nH) = \sum_{i = 1}^N \left(\frac{\partial (nH)}{\partial n_i}\right)_{P,T,n_{j\neq i}}dn_i $$

Note that for a chemical reaction $dn_i$ for each component can be related to the others by stoichiometry. For example, for a simple $\ce{A -> B}$ reaction, $-dn_A = dn_B = d\xi$, where $\xi$ is the extent of reaction. Thus, for this example,

$$ d(nH) = \left(\frac{\partial (nH)}{\partial n_A}\right)_{P,T,n_B}dn_A + \left(\frac{\partial (nH)}{\partial n_B}\right)_{P,T,n_A}dn_B = \left[-\left(\frac{\partial (nH)}{\partial n_A}\right)_{P,T,n_B}+\left(\frac{\partial (nH)}{\partial n_B}\right)_{P,T,n_A}\right]d\xi $$

Expressing this as a total derivative,

$$ \frac{d(nH)}{d\xi} = -\left(\frac{\partial (nH)}{\partial n_A}\right)_{P,T,n_B}+\left(\frac{\partial (nH)}{\partial n_B}\right)_{P,T,n_A} = \text{heat of reaction} $$

Note that the RHS is the difference between the partial molar enthalpy of each component. In ideal solutions, these correspond to the molar enthalpies of the components individually. In other words, we get the expected formula that the heat of reaction is the difference between enthalpy of products and enthalpy of reactants (weighed by stoichiometric coefficients).

So your heat of reaction is still there, you have a mixture with variable composition (a chemical reaction) therefore the enthalpy differential has an extra term related to the chemical composition.


[OP] But according to the equation $dH=nC_{p}dT$, if temperature $T$ is constant then $dT =0$ and $dH =0$.

Even if this equation would describe a system completely, a constant temperature does not imply that $dH$ is zero. Instead, you can have a temperature bath of infinite size (n approaches infinity) so that the average temperature does not increase. Of course, there would be temperature fluctuations while the reaction goes on, even if it is a glacially slow reaction.

[OP] Clearly, there are many reactions which take place at constant temperature and have a definite enthalpy change.

From the discussion above, it might be clear that it is not so easy to run an exothermic reaction at constant temperature. Instead of having a huge water bath, you can also buy an instrument for isothermal calorimetry (ITC) that monitors the temperature and cools or heats the reaction mixture to keep it at constant temperature. In this instrument, the amount of heat tranfered is logged, and that is how you measure the enthalpy change. This is distinct from measuring with a traditional calorimeter, where you monitor the temperature change and estimate the heat capacity of the calorimeter. Estimating the heat capacity is difficult because it is affected by the reactants and products you are measuring.

What this means is that there are multiple sources and sinks of thermal energy. If you are successful in balancing them perfectly, it will be an isothermal process. Often in thermodynamics, you are making arguments that rely on an ideal system. The real systems might approach that ideal, so the observation will also approach that of the ideal system.

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    $\begingroup$ OP might have meant in a theoretical sense, not practically measuring. $\endgroup$ Commented Apr 16, 2021 at 12:39
  • $\begingroup$ @SafdarFaisal Yes, they might have. Maybe they will be able to choose their favorite answer (or most useful one) once there are some more. Often, theory and measurement are related to each other, though. $\endgroup$
    – Karsten
    Commented Apr 16, 2021 at 13:04
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    $\begingroup$ This really doesn't address the main question of why the described application of the equation is invalid. $\endgroup$
    – Andrew
    Commented Apr 16, 2021 at 14:51
  • $\begingroup$ @Andrew I edited my answer, trying to address your concern while attempting to keep the approach different from your answer, which is a good one. $\endgroup$
    – Karsten
    Commented Apr 16, 2021 at 15:56

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