Help understanding delta U (or H) of reaction

[RobNik]

Hello,

I'm trying to understand the concepts of the change in internal energy (U) or enthalpy (H) of a reaction, given the laws and equations of basic thermodynamics, but I'm getting confused with the following thought experiment. I'm looking at U, since I find it easier to imagine, even though H is used more often. (Sorry but this latex entry thing seems busted, so I'll write out "delta U".) Imagine a reaction:

A + B --> C + D

has delta U = -500 kJ. Say they are all gases, and I put them in a constant volume, insulated container, so the process is adiabatic, and no work is done. (The reaction could be started with a tiny trigger, like a spark).

q = 0 [adiabatic, insulated container]
w = 0 [constant volume]
delta U = w + q = 0 + 0 = 0

Obviously the reaction is exothermic, and the temperature of this container would rise, but what exactly is the "system" for which the delta U = -500 kJ. The box as a whole has 0 delta U, right?

thanks,
Rob

 

[johng23]

Don't take my word 100% on this, but here's what I think the explanation is:

I think it's because for multi-component systems, the change in internal energy is actually not just due to work and heat, but also due to chemical changes. The fact that you say you can start this reaction with a spark implies that there is some chemical driving force for it to occur, and that driving force is the chemical potential. The expression for the change in internal energy is then

dU = TdS - pdV + $$\Sigma$$$$\mu$$$$_{i}$$dN$$_{}i$$

So in this case, the 500 kJ change in internal energy would come from converting N moles of A and B to species of lower chemical potential.

 

[DrDu]

No, U really doesn't change. There is no contradiction between the expressions of U in terms of w and q on one hand and T, S, p, V, mu and N on the other, because the reaction is irreversible, hence Delta S is greater 0 although q vanishes.

 

[johng23]

Ok, then how does that work with the equation that I posted? Does the change in the entropy term balance the change in the chemical potential term to make the net change in U zero?. Otherwise, what is the answer to the original question? How can the reaction proceed if it requires a change in U?

 

[DrDu]

Delta U is the heat that would given off to the environment if you would allow for heat exchange so that the initial and final state have the same temperature. So it does not apply to this adiabatic setting.
To be more precise: The values of Delta U which you can look up are $$\sum_i \nu_i \left .\frac{dU}{dN_i}\right|_{V,T}$$. This may be a helpful quantity to have for calculations even in adiabatic reactions, but it does not mean that it corresponds to the actual change in U which is 0.

 

[johng23]

That doesn't seem right; the control variables for U are S and V. The equation you gave is the equation for chemical potential, except it should be taken at constant S and V for internal energy. Constant T and V would correspond to Helmholtz free energy, in which case I would agree that the function does not correspond to the adiabatic situation we're discussing here. But internal energy should; an adiabatic, fixed container can correspond to constant S and V.

 

[DrDu]

So, what do you think is the correct definition of Delta U?

 

[johng23]

To be honest, I guess I don't know. I've only seen tabulated $$\Delta$$G values; in thermodynamics for solids/materials, internal energy is always introduced in the development of the theory, but only the Gibbs free energy is used in practice.

I know that $$\Delta$$G = $$\Delta$$H - T$$\Delta$$S is the tabulated value for the Gibbs energy, where H and S are calculated from the known heat capacities of the substances involved and the standard states, etc. I would think it would be somehow analagous to that. Either way, I just don't see the logic in the equation that you posted. If that really is accurate, can you explain? Why the chosen fixed variables of V and T? Don't you agree that U is not the governing potential in situations where those state variables are constant? Certainly it is still valid to evaluate a change in U in that situation, but I just don't see the reason for using U at all. In that case why not just use F?

 

[Ygggdrasil]

Hello,

I'm trying to understand the concepts of the change in internal energy (U) or enthalpy (H) of a reaction, given the laws and equations of basic thermodynamics, but I'm getting confused with the following thought experiment. I'm looking at U, since I find it easier to imagine, even though H is used more often. (Sorry but this latex entry thing seems busted, so I'll write out "delta U".) Imagine a reaction:

A + B --> C + D

has delta U = -500 kJ. Say they are all gases, and I put them in a constant volume, insulated container, so the process is adiabatic, and no work is done. (The reaction could be started with a tiny trigger, like a spark).

q = 0 [adiabatic, insulated container]
w = 0 [constant volume]
delta U = w + q = 0 + 0 = 0

Obviously the reaction is exothermic, and the temperature of this container would rise, but what exactly is the "system" for which the delta U = -500 kJ. The box as a whole has 0 delta U, right?

thanks,
Rob

Under conditions of constant volume, ΔU is equal to the amount of heat released by the reaction. Of course, if A and B release 500kJ/mol when reacting, their potential energy must be reduced by 500kJ/mol when they are converted to C and D.

Consider a simpler case of a small amount of A and B in a container filled with an inert gas like nitrogen (N₂). As A and B react, they release their energy via collision with the N₂ molecules. These collisions speed up the N₂ molecules, increasing the temperature (average kinetic energy) of the gas. Therefore, what has occurred is the excess chemical potential energy of the A and B molecules gets transferred to the kinetic energy of the N₂ molecules (which in turn gets transferred to kinetic energy of A, B, C, and D when the N₂ molecules collide with them). Therefore, the overall internal energy of the system remains constant. Energy has just been converted from chemical potential energy to kinetic energy. The increase in temperature of the system is offset by the lower potential energy of the products.

You can think of internal energy as having a thermal component (representing the molecules' kinetic energy) and a chemical component (representing the potential energy of the molecules in the system). In the reaction, we are essentially shifting energy from the chemical component to the thermal component. Under these conditions, the molecules can still lose potential energy but still have the same amount of internal energy.

 

[DrDu]

Well, Delta U can be used to calculate the change of Delta F with temperature (van't Hoff equation). But you are right it is not so usefull for an adiabatical reaction.
I checked it's definition in Prigogine, Defay, Chemical Thermodynamics and it is in deed defined at fixed T and V, as I said. Prigogine got the Nobel prize for his contributions to chemical thermodynamics, so I suppose that is authoritative.
Anyhow, it should be clear that Delta U is not in any case equal to the change in intenal energy, only for the case of an isothermal isochoric reaction, and that was the main confusion in the first post.

 

[freierz]

It's not really as complicated as some of the posters act. Thermodynamics is an elegant subject, because there are so many questions that have simple, cut and dry answers if you look at them the right way.

First of all, U IS in fact internal energy. It represents the sum of the energy of the system (but since there's no such thing as 0 energy, we usually discuss delta U).

Second, a simple application of the first law of thermodynamics says, if the system is adiabatic, delta U = 0. Period. It really is no more complicated than that. If no energy is being transferred between the system and surroundings, the system will neither increase nor decrease in energy.

In your example, where there is a reaction, the relevant quantity is the enthalpy of reaction (delta H). The energy balance equation is

delta U = Q + W + delta H + (integral)Cp*dT

This accounts for heat transfer with the surroundings (Q), expansion/contraction work (W), the reaction (delta H) and the temperature of the system (last term). Since we have a fixed container and it's adiabatic, Q and W are both 0. delta U is 0 as well, so in conclusion, the change in energy due to reaction causes a change in temperature.

 

[juanrga]

Hello,

I'm trying to understand the concepts of the change in internal energy (U) or enthalpy (H) of a reaction, given the laws and equations of basic thermodynamics, but I'm getting confused with the following thought experiment. I'm looking at U, since I find it easier to imagine, even though H is used more often. (Sorry but this latex entry thing seems busted, so I'll write out "delta U".) Imagine a reaction:

A + B --> C + D

has delta U = -500 kJ. Say they are all gases, and I put them in a constant volume, insulated container, so the process is adiabatic, and no work is done. (The reaction could be started with a tiny trigger, like a spark).

q = 0 [adiabatic, insulated container]
w = 0 [constant volume]
delta U = w + q = 0 + 0 = 0

Obviously the reaction is exothermic, and the temperature of this container would rise, but what exactly is the "system" for which the delta U = -500 kJ. The box as a whole has 0 delta U, right?

thanks,
Rob

$U_{box} = U_{molecules} + K_{molecules} + ···$

The internal energy of the box is due to the linear motion of the molecules, more the energy of rotations, more the internal energy of the molecules...