Appunti Chemical reaction engineering

Chemical reaction engineering

Lez. 1 - 30/09/2024

Review

NC (N° of components) NR (N° of reactions) NP (N° of phases)

• Pure specie 1 0 1
• 1 0 >1
• Homogeneous >1 0 1 mixture
• Process separat. >1 1 1
• Subject >1 >1 1

We measure the amount of matter in a substance using volume or mass. When studying chemical reactions, we often refer to molecules. However, since the number of molecules involved is typically enormous and impossible to count individually, we use the concept of moles. A mole is defined as 6.022×10 molecules, known as Avogadro’s number. 23

Moles are related to mass and volume through parameters such as atomic weight and molecular weight.

Lez. 2 - 01/10/2024

Pure species

NC=1 NR=0 NP=1

Yesterday, we discussed single species in a single phase, focusing on measuring the amount. We can do this using:

• m : [ g, mg, t, … ]i
• N : [ moles ]i

We can determine the number of moles from the mass of the species using:

• Molecules: m =MW Ni i i
• Atoms: m =AW Ni i i

To measure the occupied space, we can use volume, which changes with variations in temperature (T), pressure (P), or amount:

• V (T, P, amount) : [ m³, L, cm³, … ]i

Since volume is an extensive property, it is proportional to the amount. This makes direct comparisons difficult; hence, we usually base our calculations on intensive properties. For example, we can divide the volume by the amount. Depending on the measure of amount, we can obtain:

• Massive property;=
• Molar property.=

In both cases, these units depend only on temperature and pressure, unless the substance is a liquid or solid. In such cases, we can assume as a constant. For gases, however, temperature and pressure will affect molar volume.

This dependence is described in the EoS (Equations of State), there are two categories:

• Ideal gas EoS: a good model based on the approximation that molecules have no volume and do not interact, thus limiting its application. The equation is given by:. PV =N R T or =(R T)/P.i i g g
• Real gas EoS: here, we introduce approximations to account for the reality of matter. Depending on the chosen model, we can obtain cubic, virial, van der Waals, or corresponding state theories.

The compressibility factor Z should be compared to 1:

• If Z<1: the gas is more compressed than an ideal gas.
• If Z>1: the gas occupies more volume than an ideal gas.

=

We will focus only on real gas equations moving forward.

Pure species with more than 1 phase

NC=1 NR=0 NP>1

We start to think watching the P-T diagram:

In the subcritical domain, moving between two points can lead to two very different situations. In fact, the density of a liquid is typically three orders of magnitude greater than that of the gas.

Lez. 3 - 02/10/2024

Starting from information that:

• =
• = for ideal gas =

We see that molar volume depends only on the condition of the system, this for every specie that behaves as ideal gas. While the massive volume depends on the type of species involved, in fact it contains the MW in the equation:

• =

For example 1 kg of He is larger than 1 kg of N2, but 1 moles of He is equal to 1 moles of N2 if both are considered as ideal gasses.

Now returning to the P-T diagram, if we know the position of the line in the plot we could know the state of the specie, the line indicates the equilibrium between the two phases. We need to identify the T and P who permits the equilibrium between the two phases. This equilibrium is given by:, with alpha and beta two phases, G is in terms of ext (means the unit is energy) Gribbs free = energy, these are function of T,P. So we need to elaborate this dependence in order to find the saturation lines.

For doing that we know that the slope of the line is given by the Clapeyron equation:

→ ∆| = → ∆

To calculate the variation of entropy we know that: →∆→∆ = Then there is a suggestion by CLAUSIUS, in the transition from two phases the molar volume of a gas phase is much larger than the condensed phase. So the same amount of the same specie change volume due to the different phase. In fact: where L or S phase ≫ =to

The order of this difference is on the magnitude of 10. So for example 1m³ of liquid H2O weight 1 ton beside the same volume of gas H2O weight 1 kg.

Knowing that we could approximate the difference between the volume as the volume of the gas phase: →∆ ≅ ∆ = Replacing that in Clapeyron equation we get:

→ → → ∆ 1 ∆ ∆.| = ∙ = ∙ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ | = ⎯⎯⎯⎯⎯⎯⎯ → ∆ 1 1≅ ( − )

Were with 0 we indicate a reference measure. We typically take as atmospheric and find the of evaporation. So if we know , and we can find all the saturation line.→∆

The same idea could be applied to all the other lines in the diagram.

Example: If we can confirm that than we confirm that is P increases, then T boiling| − > 0 ∆ increases. To do that we know according to CLAUSIUS that , because T is in Kelvin the = denominator is always positive, and because means difference between enthalpy of gas ∆ and liquid (usually positive) we can confirm that.

“Thermal state” means a state where we are able to determine the thermal properties connected to T, that are usually H and U, this can be intensive or extensive. We want to evaluate the H( T, P ) numerically, typical thermodynamics approach consists in splitting the equation in two fixing a parameter and looking at the other: = | + | ( )

With , where for liquid or solids internal energy depends| = | = | ++ | only on temperature and are almost incompressible, for this reason the equation become | =

This term is almost 0 in most cases (this approximation works in systems where volume| doesn’t change due to variation): this is valid for liquid or solid systems, in fact in these cases the enthalpy is not significantly affected by the pressure, so This approximation is valid(, ) ≅ (). also for IG because enthalpy of an ideal gas is independent