If we consider adiabatic reactor (one which doesn't exchange energy and matter with surroundings) with some reaction in liquid phase, energy balance for such reactor is written as heat of reaction being equal to time derivative of enthalpy at any point in time. Since reaction is in liquid phase during reaction there are negligible changes in volume and pressure of reactor so that change in enthalpy is approximately equal to change in internal energy. Problem here is that how can internal energy (or enthalpy in this case) change in adiabatic reactor since it doesn't exchange energy with surroundings? Internal energy is a sum of all microscopic kinetic energy and potential energy of interaction of all bonds in the system. During chemical reaction interatomic potential energy is converted to microscopic kinetic energy (or vice versa) because of which temperature changes, but change in each type of energy must be equal due to energy conservation. Since this is the case internal energy doesn't change after reaction in adiabatic reactor. If so, why does equation of energy balance say it does change as temperature of system is changed during reaction?


2 Answers 2


Here is how it is properly supposed to play out for the reaction in your video:

Assuming an ideal solution, the enthalpy of the reactor contents can be represented as: $$H=n_Ah_a+n_Bh_B$$ where the n's are the numbers of moles of the two species and the h's are the enthalpies of the pure species at the same temperature as the reactor mixture: $$h_A(T)=h_A(T^0)+ C_{PA}(T-T^0)$$ $$h_B(T)=h_B(T^0)+ C_{PB}(T-T^0)$$where $T_0$ is a reference temperature typically taken as 298 K and $h(T^0)$ is so-called heat of formation of the species (from the pure elements at $T^0$). Taking the derivative of the reaction mixture enthalpy with respect to time yields: $$\frac{dH}{dt}=n_A\frac{dh_A}{dt}+n_B\frac{dh_B}{dt}+h_A\frac{dn_A}{dt}+h_B\frac{dn_B}{dt}$$But, $$n_A\frac{dh_A}{dt}+n_B\frac{dh_B}{dt}=(n_AC_{PA}+n_BC_{PB})\frac{dT}{dt}$$and$$\frac{dn_B}{dt}=-\frac{dn_A}{dt}=r$$where r is the rate of reaction. Therefore, $$\frac{dH}{dt}=(n_AC_{PA}+n_BC_{PB})\frac{dT}{dt}+r(h_B(T)-h_A(T))$$But, from Hess' law, the heat of reaction at temperature T is $$\Delta H_r(T)=h_B(T)-h_A(T)=h_B(T^0)-h_A(T_0)+(C_{PB}-C_{PA})(T-T^0)$$Therefore, for the reaction mixture, $$\frac{dH}{dt}=(n_AC_{PA}+n_BC_{PB})\frac{dT}{dt}+r\Delta H_r(T)$$If, for an adiabatic reactor, we set this equal to zero, we obtain: $$0=(n_AC_{PA}+n_BC_{PB})\frac{dT}{dt}+r\Delta H_r(T)$$or$$(n_AC_{PA}+n_BC_{PB})\frac{dT}{dt}=r(-\Delta H_r(T))$$This is basically the same equation as in the video

  • $\begingroup$ Thank you. I would like if people would derive energy balances for reactors in more detail (like you did here) and not just write end equation since it can create misunderstandings like it did for me. I have only one question what is connection between enthalpy of A or B and enthalpy of formation of A or B? Since enthalpy of formation is heat which is exchanged with surroundings for formation of compound from elements at constant T and P. What does such heat have to do with enthalpy of component A or B at some temperature? $\endgroup$ May 1, 2021 at 17:01
  • $\begingroup$ Basically, why is enthalpy of component A or B at reference temperature equal to enthalpy of formation of that component at same temperature? $\endgroup$ May 1, 2021 at 17:12
  • $\begingroup$ That’s the definition of heat of formation. $\endgroup$ May 1, 2021 at 17:16
  • $\begingroup$ Yes, that is what I don't quite understand. Enthalpy of formation is heat which is exchanged with surroundings at constant temperature and pressure when compound is created from elements in most stable state. Enthalpy of component is sum of U + PV of component at some temperature relative to some reference. Why would this two quantities be equal? $\endgroup$ May 1, 2021 at 17:55
  • $\begingroup$ Sorry, I can't teach an entire course in thermo here. Your background right not is too limited for me to spend the necessary time. What I hope is that, even without understand that, you still get the gist of what I was trying to show in this derivation. $\endgroup$ May 1, 2021 at 19:14

You omitted the energenics associated with breaking chemical bonds and making new chemical bonds. If this contributes a negative change to the internal energy (exothermic reaction), then the temperature would have to rise to offset this effect so that the overall change in internal energy would be zero. If it contributes a positive change to the internal energy (endothermic reaction), then the temperature would have to drop to offset this effect so that the overall change in internal energy would be zero. Saying this in a different way, when you have a chemical reaction occurring, the internal energy can change even if the temperature is held constant.

  • $\begingroup$ Yes I agree, but if internal energy change is zero in our case why does every book in which I was looking say that it changes in adiabatic reactor? Change in internal energy was used to calculate temperature difference. As I said energy balance says that heat of reaction equals time derivative of enthalpy (or internal energy in this case) at every point in time. Ths video from LearnChemE channel from Colorado University is only one example (you can skip to the part where he writes energy balance for adiabatic reactor): youtu.be/--cZwa2TXhY $\endgroup$ May 1, 2021 at 8:04
  • $\begingroup$ I don't know what books you are using, but, if they say that, they are incorrect. In your video, the lecturer is incorrect with regard to the sign of the heat of reaction on the right hand side of the equation. It should be a minus sign (see Fogler's book on reactor design). If you move that term to the other side of the equation, it is going to start looking a lot more like dH/dt = 0 for the reaction mixture (where they are assuming that the reaction is occurring at constant pressure). $\endgroup$ May 1, 2021 at 11:30
  • $\begingroup$ Regarding reaction enthalpy in the video, I think minus sign denotes the fact that exothermic reactions ,which tend to increase temperature of reactor, have negative enthalpy of reaction - nothing too important. But, issue I have is that he says enthalpy changes in adiabatic reactor with temperature. Derivative of enthalpy with time is different from zero. How can this be since we agree that enthalpy (or internal energy in this case) can't change in such reactor? $\endgroup$ May 1, 2021 at 12:56
  • $\begingroup$ Who says that he says that? If he does, he is wrong. The derivative of the enthalpy with respect to time is zero. That's basically what his equation amounts to. $\endgroup$ May 1, 2021 at 13:35
  • $\begingroup$ Did you see the equation he wrote? He connected time derivative of enthalpy with change of with emperature (its time derivative). dH/dt = [Nacp(a) + Nbcp(b)]*dT/dt = heat of reaction $\endgroup$ May 1, 2021 at 13:42

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