# Change of P and T after a chemical reaction

Gold Member
If I have a chemical reaction in closed system, how can I evaluate the change in T and P?

I have:
- one equation for the chemical equilibrium
- one equation for the energy balance

but I need an other one! entropy balance, maybe?

Thanks
Ric

BvU
Homework Helper
How about an equation of state ? And phase equilibrium - if more phases are present.

dRic2
Gold Member
So simple! Thank you!!

Anyway, if more phases are present (and of course I have to include the equations for phase equilibria) how do I set up the EoS? Should I just choose a random component of the system and write the EoS for it ?

BvU
Homework Helper
and of course I have to include the equations for phase equilibria
well, I mentioned them because you did not

Is this homework ? What is the context for this exercise ? You have a library full of books on physical properties for chemistry -- there must be a section on mixing rules in one of them ?

Gold Member
No I was just thinking. When I studied thermodynamic usually I ran into exercises with constant P but I never had the chance to study a system with both a change in T and P

PS: I totally forgot about mixutres, thanks again! ;)

Chestermiller
Mentor
Would you know how to do it if there were no phase change and you were dealing with an ideal gas mixture?

Gold Member
Yes

Gold Member
It would be something like this (I guess):

##dG_{T, P} = 0## (chemical equilibrium)
##ΔU = Q + L → ΔU = 0##
##pv=RT##

Chestermiller
Mentor
It would be something like this (I guess):

##dG_{T, P} = 0## (chemical equilibrium)
##ΔU = Q + L → ΔU = 0##
##pv=RT##
##\Delta U=0## would have to be accompanied by some consideration of the standard internal energy change for the reaction and the change in sensible heat of the products. There is an analysis of this in a recent thread. I'll try to locate the link.

Chet

Gold Member
ΔU=0ΔU=0\Delta U=0 would have to be accompanied by some consideration of the standard internal energy change for the reaction and the change in sensible heat of the products.

I would do like I do if I have to work with enthalpy:

##U_A = U_B##

##U_A = \sum n_{i, A}( u_i(T_r) + \int_{T_r}^{T_A} c_{v, i} dT)##

where ##T_r## is an arbitrary temperature (of reference).

##U_B = \sum n_{i, B} (u_i(T_r) + \int_{T_r}^{T_B} c_{v, i} dT) = \sum (n_{i, A} + \lambda*v_i)(u_i(T_r) + \int_{T_r}^{T_A} c_{v, i} dT + \int_{T_A}^{T_B} c_{v, i} dT ) ##

where ##\lambda## is the extent of reaction and ##v_i## is the stoichiometric coefficient for the i-specie. I do the product and I get:

##U_B = \sum n_{i, A} (u_i(T_r) + \int_{T_r}^{T_A} c_{v, i} dT) + \sum (n_{i, A} \int_{T_A}^{T_B} c_{v, i} dT ) + \sum \lambda*v_i * (u_i(T_r) + \int_{T_r}^{T_A} c_{v, i} dT + \int_{T_A}^{T_B} c_{v, i} dT )##

##U_B = \sum n_{i, A} (u_i(T_r) + \int_{T_r}^{T_A} c_{v, i} dT) + \sum (n_{i, A} \int_{T_A}^{T_B} c_{v, i} dT ) + \sum \lambda*v_i * (u_i(T_r) + \int_{T_r}^{T_B} c_{v,i} dT)##

So ##U_A-U_B## is:

##U_A-U_B = \sum (n_{i, A} \int_{T_A}^{T_B} c_{v, i} dT ) + \sum \lambda*v_i * (u_i(T_r) + \int_{T_r}^{T_B} c_{v,i} dT) = 0##

Extracting ##\lambda## from the summation I get:

## \sum (n_{i, A} \int_{T_A}^{T_B} c_{v, i} dT ) = - \lambda \sum v_i * (u_i(T_r) + \int_{T_r}^{T_B} c_{v,i} dT)##

Now I set ##u(T_r) + \int_{T_r}^{T_B} c_{v,i} dT = Δu_{formation, i}(T)## and so ##\sum v_i*Δu(T)_{formation, i} = ΔU(T)_{reaction}## just like I would do with ##h##.

Is it wrong ?

Last edited:
Chestermiller
Mentor
I would do like I do if I have to work with entalpy:

##U_A = U_B##

##U_A = \sum n_{i, A}( u_i(T_r) + \int_{T_r}^{T_A} c_{v, i} dT)##

where ##T_r## is an arbitrary temperature (of reference).

##U_B = \sum n_{i, B} (u_i(T_r) + \int_{T_r}^{T_B} c_{v, i} dT) = \sum (n_{i, A} + \lambda*v_i)(u_i(T_r) + \int_{T_r}^{T_A} c_{v, i} dT + \int_{T_A}^{T_B} c_{v, i} dT ) ##

where ##\lambda## is the extent of reaction and ##v_i## is the stoichiometric coefficient for the i-specie. I do the product and I get:

##U_B = \sum n_{i, A} (u_i(T_r) + \int_{T_r}^{T_A} c_{v, i} dT) + \sum (n_{i, A} \int_{T_A}^{T_B} c_{v, i} dT ) + \sum \lambda*v_i * (u_i(T_r) + \int_{T_r}^{T_A} c_{v, i} dT + \int_{T_A}^{T_B} c_{v, i} dT )##

##U_B = \sum n_{i, A} (u_i(T_r) + \int_{T_r}^{T_A} c_{v, i} dT) + \sum (n_{i, A} \int_{T_A}^{T_B} c_{v, i} dT ) + \sum \lambda*v_i * (u_i(T_r) + \int_{T_r}^{T_B} c_{v,i} dT)##

So ##U_A-U_B## is:

##U_A-U_B = \sum (n_{i, A} \int_{T_A}^{T_B} c_{v, i} dT ) + \sum \lambda*v_i * (u_i(T_r) + \int_{T_r}^{T_B} c_{v,i} dT) = 0##

Extracting ##\lambda## from the summation I get:

## \sum (n_{i, A} \int_{T_A}^{T_B} c_{v, i} dT ) = - \lambda \sum v_i * (u_i(T_r) + \int_{T_r}^{T_B} c_{v,i} dT)##

Now I set ##u(T_r) + \int_{T_r}^{T_B} c_{v,i} dT = Δu_{formation, i}(T)## and so ##\sum v_i*Δu(T)_{formation, i} = Δu(T)_{reaction}## just like I would do with ##h##.

Is it wrong ?
It's OK as long as the number of moles of gas doesn't change between the reactants and the products. In that case, the standard change in internal energy (at constant volume) would be equal to the standard change in enthalpy (at constant pressure). However, if there is a change in the number of moles, the standard internal energy change of reaction and the standard enthalpy change differ.

Gold Member
I don't get it... My reasoning seems to me the same whether the number of moles changes or not. In my previous post U refers to internal energy, not enthalpy.

Chestermiller
Mentor
I don't get it... My reasoning seems to me the same whether the number of moles changes or not. In my previous post U refers to internal energy, not enthalpy.
Sorry. I got confused when you mentioned enthalpy. In addition, the data you are going to have available is almost certainly going to be only enthalpies of formation, not internal energies of formation. So this has to be converted to standard internal energies of the reaction, and the conversion is going to involve the change in the number of moles.

Gold Member
Sorry. I got confused when you mentioned enthalpy.

I'm a pretty messy guy in writing, my fault. Anyway I'd like to know if that is correct because I've never encountered an internal energy of formation. I don't think it is a big deal, but I'd like to be sure.

Chestermiller
Mentor
I'm a pretty messy guy in writing, my fault. Anyway I'd like to know if that is correct because I've never encountered an internal energy of formation. I don't think it is a big deal, but I'd like to be sure.
I think what you did is valid. Basically, what you are using are the internal energies at the standard temperature and pressure.

Gold Member
In addition, the data you are going to have available is almost certainly going to be only enthalpies of formation, not internal energies of formation. So this has to be converted to standard internal energies of the reaction, and the conversion is going to involve the change in the number of moles.

I thought internal energy of reaction is simply - by definition - ##\Delta U_R(T) = \sum v_i \Delta u_{f, i}^°(T)## where ##U_{f, i}^°(T)## is the internal energy of formation of i. (The number of moles does not play a role here, but I may be wrong)

So the real problem, as you pointed out, is how to evaluate ##u_{f, i}^°(T)##.

Chestermiller
Mentor
I thought internal energy of reaction is simply - by definition - ##\Delta u_R(T) = \sum v_i \Delta u_{f, i}^°(T)## where ##u_{f, i}^°(T)## is the internal energy of formation of i. (The number of moles does not play a role here, but I may be wrong)

So the real problem, as you pointed out, is how to evaluate ##u_{f, i}^°(T)##.
As I said, you are not going to find internal energies of formation in tables. You are going to find enthalpies of formation in tables. And to convert to internal energies of formation, you are going to need a ## Pv## for each species.

dRic2
Gold Member
Thank you, I have to study this topic a little bit more because I do not remember it very well.