Internal Energy Changes in Adiabatic Reactor

Question

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?

Answer 1

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

Answer 2

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.