Industrial Chemical Processes - Notes, Appunti di Ingegneria dei Processi di Produzione Industriale

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INDUSTRIAL

CHEMICAL

PROCESSES

STRUCTURE of the CHEMICAL INDUSTRY

The chemical industry processes raw materials (organic and inorganic) and converts them into substances and products according to the needs of consumers or other industries. Chemical products may have many applications and uses: ● raw or basic materials for other industries ● intermediate products for other industries ● finished products for the industrial system, agriculture, services and consumption

THERMODYNAMICS

Thermodynamics gives information about: ● the possibility of a chemical reaction to occur and to what extent it occurs ● The effect of temperature and pressure on the equilibrium (of a chemical reaction) ● The composition of reactor effluents if equilibrium is reached ● The driving force of each of several competing reactions taking place ● The amount of heat released or absorbed during a reaction but it can’t tell anything about: ● the rate of the reaction ● the effect of type and shape of the reactor used ● the relative rates of competing reactions. Gibbs free energy for homogeneous systems is defined by two chemical-physical variables

and by its composition: 𝐺 = 𝐺 𝑇, 𝑝, 𝑛( 1 ,..., 𝑛𝑖).

If the general expression of Gibbs free energy is differentiated:

𝑑𝐺 = (^) ( ∂𝐺∂𝑇) 𝑝,𝑛𝑖

𝑑𝑇 + (^) ( ∂𝐺∂𝑝) 𝑇,𝑛𝑖

𝑖

( (^) 𝑖) 𝑇,𝑝,,𝑛𝑗≠𝑖

Considering an isothermal and isobaric transformation 𝑑𝐺 =. 𝑖

( (^) 𝑖) 𝑇,𝑝,,𝑛𝑗≠𝑖

The chemical potential of the species i is defined as

μ𝑖 = (^) ∂𝑛∂𝐺 ( (^) 𝑖)𝑇,𝑝,,𝑛 𝑗≠𝑖 After defining the chemical potential, Gibbs free energy can be expressed as

𝑑𝐺 = (1) 𝑖

∑ μ𝑖𝑑𝑛𝑖 = 𝑖

∑ ν𝑖μ𝑖𝑑ξ

where: ● ν𝑖is the stoichiometric coefficient of the i-th species ● ξis the extent of reaction From (1), an expression for the variation of Gibbs free energy can be obtained:

∂𝐺 ( (^) ∂ξ)𝑇,𝑝^ =^ ∑ ν𝑖 𝑖μ𝑖 = ∆𝐺

The equilibrium constant can be expressed as 𝐾𝑎 = 𝐾𝑦𝐾φπ 𝑖

∑ν𝑖

where y are the molar fractions of the species, φ are their fugacity coefficients and πis the total pressure. If we are considering an ideal reaction system, 𝐾φ = 1, so the equilibrium constant for the

generic reaction 𝑎𝐴 + 𝑏𝐵 ↔ 𝑟𝑅 + 𝑠𝑆becomes

𝑦𝑅𝑟^ 𝑦𝑆𝑆 𝑦𝐴𝑎𝑦𝐵𝑏^

𝑟+𝑠−𝑎−𝑏

HETEROGENEOUS SYSTEMS

In heterogeneous systems , composed by different phases, a reaction can take place if its

variation of Gibbs free energy is ∆𝐺 = α

𝑁 𝑝ℎ𝑎𝑠𝑒𝑠 ∑ 𝑖

𝑁 𝑠𝑝𝑒𝑐𝑖𝑒𝑠 ∑ ν𝑖αμ𝑖α^ < 0

where μ𝑖are the chemical potential of species i in the phase. α α

The reaction is at chemical equilibrium if ∆𝐺 = and this is possible α

𝑁 𝑝ℎ𝑎𝑠𝑒𝑠 ∑ 𝑖

𝑁 𝑠𝑝𝑒𝑐𝑖𝑒𝑠 ∑ ν𝑖αμ𝑖α^ = 0

only if the chemical potential of the species in different phases are equal :

μ𝑖α^ = μ𝑖β^ =... = μ𝑖γ

APPLICATION to EQUILIBRIUM

Let’s consider the equimolar ideal gas reaction 𝐴 + 𝐵 ↔ 𝑅 + 𝑆. To find the molar fraction of a component using the equilibrium constant we can set the initial molar fraction of A and B at 𝑥 0. Considering the stoichiometry of the reaction, at time t the molar fraction of the

system will be: ● 𝑥𝐴 = 𝑥𝐵 = 𝑥 0 − 𝑥 ● 𝑥𝑅 = 𝑥𝑆 = 𝑥

and substituting this inside the expression of ∆𝐺: ∆𝐺 = ∆𝐺 → 0

• 𝑅𝑇 𝑖

0

• 𝑅𝑇 𝑖

𝑥𝑅^1 𝑥𝑆^1 𝑥𝐴^1 𝑥𝐵^1 π 1+1−1− ∆𝐺 = ∆𝐺 0

• 𝑅𝑇 𝑖

2

(^ 𝑥 0 −𝑥)^2

With this expression it’s possible to graph the trend of ∆𝐺at constant temperature. Using the limit trends, it’s possible to see that, when 𝑥 → 0 then ∆𝐺 → − ∞so there is a high tendency to form products. When instead, 𝑥 → 𝑥 0 , ∆𝐺 → + ∞so the inverse reaction is favored.

At thermodynamic equilibrium ∆𝐺 = 0 so it’s possible to find the value of 𝑥 at equilibrium :

𝑥𝑒𝑞 =

𝑥 0 1+𝑒𝑥𝑝 ∆𝐺 0 ( 2𝑅𝑇)

CALCULATION of the YIELD at EQUILIBRIUM

An important application of ∆𝐺^0 (achievable, for example, using the ∆𝐺𝑓^0 of formation of the

chemical species involved in the reaction) concern the calculation of the yield at the thermodynamic equilibrium for a given industrial process. Since the position of equilibrium is defined by the equilibrium constant it’s important to remember that an high equilibrium constant means that at the equilibrium the system is made up mostly by products.

From the expression ∆𝐺^0 =− 𝑅𝑇 𝑙𝑛 𝐾𝑒𝑞 we say by convention that a reaction is

thermodynamically favored when ∆𝐺^0 < 0 and so 𝐾𝑒𝑞 > 1(in an appropriate and possibly

wide range of temperature).

Knowing ∆𝐺 it’s possible to calculate and then the actual yield. 0 𝐾𝑒𝑞

ΔG of FORMATION

● When ∆𝐺𝑓is extremely negative , the compound is more stable with respect to the 0

elements that compose it. ● Knowing ∆𝐺𝑓^0 it’s possible to compute the ∆𝐺𝑅^0 and so the equilibrium yields (at constant temperature) → this is not true if there is some material exchange with the external

SUMMARY - ΔG

The trend progressively increases less until it reaches the maximum value at thermodynamic equilibrium.

FAVORING a REACTION

THERMODYNAMICALLY

Let’s see the difference between the yield obtained for a endothermic and exothermic reaction while changing the temperature:

In the case of an endothermic reaction ( ∆𝐻^0 > 0), increasing the temperature makes the equilibrium value for yield to be higher since endothermic reactions are favored thermodynamically by higher temperatures.

In the case of an exothermic reaction ( ∆𝐻^0 < 0), increasing the temperature makes the equilibrium value for yield to be lower since these reactions are favored thermodynamically by lower temperatures. From these graphs it can be seen also that at higher temperatures , the initial slope of the trend is higher since increasing the temperature makes the reaction faster. Furthermore, generally:

● ∆𝑆^0 > 0for reaction in which there is an increase in the number of moles

● ∆𝑆 for reaction in which there is a decrease in the number of moles 0 < 0

Since ∆𝐺^0 = ∆𝐻^0 − 𝑇∆𝑆^0 from a general point of view there are 4 different situations:

● ∆𝐻 and : it’s an exothermic reaction which is highly favored → 0 < 0 ∆𝑆 0 > 0 ∆𝐺 for every temperature , furthermore the equilibrium moves towards 0 < 0 products increasing temperature ● ∆𝐻 and : it’s an exothermic reaction which is favored at low 0 < 0 ∆𝑆 0 < 0 temperatures (also because ∆𝐺 is negative if T is sufficiently low) 0 = ∆𝐻 0 − 𝑇∆𝑆 0

● ∆𝐻^0 > 0 and ∆𝑆^0 > 0: it’s an endothermic reaction which is favored by high temperatures (also because ∆𝐺 is negative if T is sufficiently high) 0 = ∆𝐻 0 − 𝑇∆𝑆 0

● ∆𝐻^0 > 0 and ∆𝑆^0 < 0: it’s an endothermic reaction which is always disadvantaged since ∆𝐺^0 > 0

It can be seen that at low temperatures, the value of ∆𝐺 is given mainly by the value of 0 ∆𝐻 0

CONVERSION and EXTENT of REACTION

The variation of the number of moles of a species i is connected with the extent of reaction ξ

→ 𝑑𝑛𝑖 = ν𝑖𝑑ξ and integrating to any point during the reaction → 𝑛𝑖,

𝑛𝑖 ∫ 𝑑𝑛𝑖 = 0

ξ ∫ ν𝑖𝑑ξ

The conversion of a reactant i is

𝑛𝑖,0−𝑛𝑖

𝑛𝑖,0^ =^

𝑛𝑖,0−𝑛𝑖,0−ν𝑖ξ

𝑛𝑖,0^ =^

ν𝑖ξ 𝑛𝑖,

from which the extent of reaction is ξ =.

𝑛𝑖,0𝑋𝑖 −ν𝑖

For irreversible reactions, the maximum value of conversion is the one of complete conversion 𝑋 = 1. For reversible reactions, the maximum value of conversion is the equilibrium conversion 𝑋 = 𝑋𝑒𝑞.

EXAMPLE

Consider the reaction 𝐶𝐻 4 + 𝐻 2 𝑂 ↔ 𝐶𝑂 + 3𝐻 2 which has ∆𝐻^0 (700°𝐶) = 53790 (^) 𝑚𝑜𝑙𝑐𝑎𝑙.

Between the range 600 𝐾 < 𝑇 < 1500 𝐾 the following expression is valid

∆𝐺𝑅^0 = 55717 − 60, 25 · 𝑇(𝐾) so ∆𝐺𝑅^0 = 0 when 𝑇 = 925 𝐾.

The reaction is operated at 925 K, 1 atm and with an equimolecular mixture of CH 4 and H 2 O. Compute the extent of reaction and the yield at equilibrium in this condition and also if the reaction is operated at 50 atm. Considering the extent of reaction as a variation in the number of moles it’s possible to compute the molar fraction of the component at equilibrium using the stoichiometry :

● 𝑦𝐶𝐻 4

= (^) 2−ξ−ξ+ξ+3ξ1−ξ = (^) 2+2ξ1−ξ

SELECTIVITY

The selectivity of a certain product R with respect to reactant A is:

𝑅

𝑛𝑅, 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑛𝐴, 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑

𝑎

𝑟 =^

𝑛𝑅−𝑛𝑅, 𝑛𝐴,0−𝑛𝐴

𝑎 𝑟

There is a correlation between conversion, yield and selectivity which is:

𝑅

𝑅

Generally, by increasing the conversion, the selectivity decreases thus it’s necessary to define

the optimal conditions to maximize the product 𝑋𝐴 · 𝑆𝐴𝑅, i.e. maximize the yield^ η𝐴𝑅.

The selectivity can be expressed also for a desired product D with respect with an undesired one U :

𝐷

𝑛𝐷, 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑

𝑛𝑈, 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑^ =^

𝑛𝐷−𝑛𝐷, 𝑛𝑈−𝑛𝑈,

EFFECTS on EQUILIBRIUM

EXAMPLE Consider the acetylene synthesis from methane 2𝐶𝐻 4 ↔ 𝐶 2 𝐻 2 + 3𝐻 2 (with

∆𝐻^0 (1500°𝐶) = 97000 (^) 𝑚𝑜𝑙𝑐𝑎𝑙 ) has ∆𝐺^0 = 95676 − 67, 2 · 𝑇(𝐾) so ∆𝐺^0 = 0 at

𝑇 = 1440 𝐾.

The reaction is operated at 1 bar with 2 initial moles of methane. At this temperature, considering ideal gases, compute the extent of reaction and the yield of methane with respect to acetylene at equilibrium. The molar fractions of the compounds at equilibrium are:

● 𝑦𝐶𝐻 4

= (^) 2−2ξ+ξ+3ξ2−2ξ = 2−2ξ2+2ξ = 1−ξ1+ξ

● 𝑦𝐶 2 𝐻 2

= (^) 2+2ξξ

● 𝑦𝐶𝐻 4

= (^) 2+2ξ3ξ

The equilibrium constant is 𝐾𝑒𝑞 = from which 𝑖

𝑁 𝑠𝑝𝑒𝑐𝑖𝑒𝑠 ∏ 𝑦𝑖

ν𝑖 ( ) · 𝑝^

𝑖 ∑ν𝑖 =

ξ ( (^) 2+2ξ) 3ξ ( (^) 2+2ξ)

3 1−ξ ( (^) 1+ξ)

2 = 1

ξ = 0, 66.

The yield of methane with respect to acetylene is η𝐶𝐻. 4

𝐶 2 𝐻 2

𝑛𝐶 2 𝐻 2 𝑛𝐶𝐻 4 ,

𝑎 𝑟 =^

ξ 2

2 1 = 0, 66

Considering a temperature of 1300 K the value of the equilibrium constant is different, in fact

𝑙𝑛 𝐾𝑒𝑞 =− ∆𝐺.

0 𝑅𝑇

∆𝐺^0 = 95676 − 67, 2 · 𝑇(𝐾) → ∆𝐺^0 = 95676 − 67, 2 · 1300 = 8316 (^) 𝑚𝑜𝑙𝑐𝑎𝑙.

𝐾𝑒𝑞 = 𝑒𝑥𝑝 − ∆𝐺.

0 ( (^) 𝑅𝑇) = 𝑒𝑥𝑝 −^

8316 ( (^) 𝑅·1300) = 0, 04

To find the value of extent of reaction 𝐾𝑒𝑞 = 𝑖

𝑁 𝑠𝑝𝑒𝑐𝑖𝑒𝑠 ∏ 𝑦𝑖

ν𝑖 ( ) · 𝑝^

𝑖

∑ν𝑖

ξ ( (^) 2+2ξ) 3ξ ( (^) 2+2ξ)

3 1−ξ ( (^) 1+ξ)

from which ξ = 0, 365 and η𝐶𝐻. 4

𝐶 2 𝐻 2

𝑛𝐶 2 𝐻 2 𝑛𝐶𝐻 4 ,

𝑎 𝑟 =^

ξ 2

2 1 = 0, 365

NB

In cases where the reaction needs heat (if they are endothermic), heating through conduction sometimes isn’t very useful because of the heat transfer so it’s necessary to generate heat directly inside the reactor. Common solutions are: ● make the methane to combust with oxygen ● electric arc ● plasma generation In this particular acetylene production, high temperature will make the product decompose since it’s highly unstable. The time spent in the reactor has to be therefore really short and a rapid cooling section is needed after it to stop the decomposition.

Since the reaction occurs with a decrease in the number of moles, it is convenient to operate at very high pressures.

EXAMPLE In the case of the synthesis of CH 3 OH , 𝐶𝑂 + 2𝐻 2 ↔ 𝐶𝐻 3 𝑂𝐻, we need to increase the

temperature to have an acceptable rate of reaction, but as low as possible to have also a high conversion (exothermic reaction). Furthermore, since there are many other products that are more stable, a selective catalyst is needed → in fact in this case, without a selective catalyst there will be a production of CH 4 instead of CH 3 OH.

EXAMPLE Considering the direct oxidation of ethylene to ethylene oxide

𝐶𝐻 2 = 𝐶𝐻 2 + 12 𝑂 2 ↔ 𝐶𝐻 2 𝐶𝐻 2 𝑂 , if the engineer doesn’t take any precaution, the product

will be the one of the total combustion (which is highly favored) 𝐶𝑂 2 and 𝐻 2 𝑂.

To avoid their formation it’s necessary a selective catalyst and a really low temperature, to unfavor the total combustion, whose products will always be present in some quantities.

EXAMPLE In the hydration of ethylene to ethanol 𝐶𝐻 2 = 𝐶𝐻 2 + 𝐻 2 𝑂 ↔ 𝐶𝐻 3 𝐶𝐻 2 𝑂𝐻

a H 3 PO 4 catalyst is used and the temperature should be higher than 300°C in order to activate it. Even at this temperature the conversion isn’t very high so it’s necessary to recycle the unreacted reactants. The variation in enthalpy is very low so the temperature control is quite easy. It’s moreover possible to increase the pressure in order to increase the equilibrium conversion.

KINETICS

The most important are the following ones: ● Evaluate the rate of reaction or the relative rates of competing reactions ● Design of industrial reactors ● Searching for the maximum yield conditions ● Automatic control of the reactor’s operation For liquid phases, we usually describe the rate of this reaction by a rate equation as shown in the following equation:

𝑛

𝑚

n+m is the total overall order of the reaction. For gas phases , the concentrations of the reacting species are often expressed as the partial pressure of the gas, and the rate of the reaction is written:

𝑚𝑜𝑙𝑒𝑠

𝑛

𝑚

The term k is called the kinetic constant which depends on temperature from the Arrhenius equation

𝑘 2

𝑘 1 =^

𝐸𝑎 𝑅

1

𝑇 1 −^

1 ( 𝑇 2 )

Its unit depends on the overall order of reaction. For more complex cases like reversible reactions we usually assign a kinetic constant to each direction :

𝑛

𝑚

𝑟

𝑠

As we describe reactions, we should be aware that the reactants for a desired reaction are also often capable of making other side reactions to occur. In other words, we are often dealing with two or more competing reactions. In such cases, the challenge is to maximize the desired reaction rate and minimize the undesired reaction rate : ● by selecting concentrations , temperatures and pressure ( process variables ) that maximize the ratio of desired-to-undesired products ● by choosing a very selective catalyst. There is a correlation between the kinetic constant and the equilibrium constant : 𝑘𝑑𝑖𝑟

𝑘𝑖𝑛𝑣^ = 𝐾𝑐,𝑒𝑞

In gas phase.

𝑘𝑑𝑖𝑟 𝑘𝑖𝑛𝑣^ · (𝑅𝑇)

− = 𝐾𝑝,𝑒𝑞

Space time τis usually applied only to flow processes, and is the time required to process one volume of feed equal to the reactor volume, measured at inlet conditions or at some standard conditions → is the time required for a volume of feed, equal to the volume of the vessel (or better the volume occupied by liquid), to flow through the vessel.

𝑉𝑅

𝑞 0 =^

𝑐𝐴,0𝑉𝑅 𝑁𝐴,

These quantities are often referred at inlet conditions

● Gas hourly space velocity 𝐺𝐻𝑆𝑉 =

𝑞 0 𝑉𝑅^ =^

𝑞 0 𝑉𝑝𝑎𝑐𝑘𝑒𝑑 𝑏𝑒𝑑

● Liquid hourly space velocity 𝐿𝐻𝑆𝑉 =

𝑞 0 𝑉𝑅^ =^

𝑞 0 𝑉𝑝𝑎𝑐𝑘𝑒𝑑 𝑏𝑒𝑑

→ low values states that an insufficient amount of feed is being processed and an higher contact time with the catalyst is needed In catalyzed reactors that employ solid catalysts, it is also used the “Weight hourly space velocity”

𝑊𝐻𝑆𝑉 =

𝐹 0 𝑚𝑐𝑎𝑡𝑎𝑙𝑦𝑠𝑡

In the case of reaction with solid catalyst (heterogeneous catalysis) that has a mass 𝑚𝑐𝑎𝑡𝑎𝑙𝑦𝑠𝑡

and with a catalytic bed packing density ρ𝐵, the apparent volume occupied by the catalyst

𝑉𝑐𝑎𝑡 shall be:𝑉𝑐𝑎𝑡 =

𝑚𝑐𝑎𝑡 ρ𝐵

The space time and the space velocity will be:

τ = and

𝑉𝑐𝑎𝑡

𝑞 0 𝑆𝑉 =^

1

τ =^

𝑞 0 𝑉𝑐𝑎𝑡

If the volumetric flowrate is compared to the weight of the catalyst instead of its volume

( τ = ), the unit of measurement of space time will be so space velocity will get

𝑚𝑐𝑎𝑡 𝑞 0

𝑘𝑔𝑐𝑎𝑡·ℎ 𝑚^3

the unit of measurement 𝑚 → this type of space time is important to compare different

3 𝑘𝑔𝑐𝑎𝑡·ℎ

catalysts in the same reactor.

NB Reactions of dehydrogenation are often endothermic while hydrogenation is exothermic.

In the case of heterogeneous catalysis, the reaction rate can be expressed relative to the

specific surface of the catalyst 𝑆 𝑚 instead of the reaction volume. This approach is useful

2 𝑔

when comparing catalysts with different surface areas:

1 𝑆

𝑑𝑛𝐴 𝑑𝑡 = 𝑘𝑓 𝑐(^ 𝐴)

From an industrial standpoint, expressing the effective rate of reaction with respect to the catalyst mass 𝑚𝑐𝑎𝑡is more useful.

1 𝑚𝑐𝑎𝑡

𝑑𝑛𝐴 𝑑𝑡 = 𝑘𝑓 𝑐(^ 𝐴)

REACTION RATE as A FUNCTION of

TEMPERATURE

EXOTHERMIC REACTION

In exothermic reversible reactions, products have lower energy than reactants and this means that the activation energy of the direct reaction is lower than the inverse. Increasing the temperature favors the direct reaction until this increase changes the characteristic of the reaction itself: the products have a higher energy than the reactants so the inverse reaction starts occurring , reaching the equilibrium faster. The temperature at which the inverse reaction is the favored one is:

𝑇𝑚𝑎𝑥 =−

∆𝐻𝑅 𝑅 𝑙𝑛 𝐾𝑒𝑞,𝑐

𝑘𝑖𝑛𝑣 𝑘𝑑𝑖𝑟

𝐸𝑎,𝑖𝑛𝑣 ( 𝐸𝑎,𝑑𝑖𝑟)

−

This can also be seen in the Arrhenius equation for the kinetic constant, which will have a higher k for the direct reaction.

INTERPRETATION OF EXPERIMENTAL DATA The equations for the rate of reaction are deduced from experimental data; in the laboratory it’s convenient to determine the reaction isotherms (yield achievable at a given T as a function of the contact time/residence time) with a small reactor maintained at constant temperature.

ISOCHRONE CURVES

Isochrone curves consist in plotting the yield of a process as a function of the temperature with parametric curves in contact time. Considering for example an exothermic reaction: ● the higher the contact time the higher the conversion (true also for endothermic) ● the higher the temperature and the faster the equilibrium is reached ● if temperature is low the achievable conversion is higher ● all the curves must be below the τ = ∞curve, which represent the equilibrium yield

ISOCONVERSION CURVES

Isoconversion curves are curves with a constant conversion of a certain reactant. ● the intersections of the isoconversion curves with the abscissa represent the equilibrium points where the reaction stops ● the reaction is exothermic → at higher temperatures the maximum conversion obtained is lower

● there is a path, represented by the dotted line, which intersects the isoconversion curves where the reaction rate is at its maximum → following this path the process gets its maximum productivity as both conversion and rate of reaction have reasonable values Maximizing the rate of reaction means decreasing the volume of the equipment and therefore τ. The aim of the engineer is to approximate the optimal temperature profile along the reactor. One solution (for exothermic reactions) is to subdivide the reactor into adiabatic catalytic layers with inter-cooling stages. In the catalytic layers there’s no heat removal, thus the temperature increases as well as the reaction rate. In the cooling section, realized by heat exchangers temperature decreases in order to approach the optimal profile.

EXAMPLE - AMMONIA SYNTHESIS

In the process of ammonia synthesis, for example:

1. Feed rises inside the wall of the reactor to be preheated
2. the gas feed enters the actual reactor near the maximum temperature at which the catalyst is active
3. it starts to go down in the catalytic bed, causing an increase in temperature (exothermic reaction) and therefore an increase in rate of reaction
4. when the maximum reaction rate is reached the mixture enters a cooling section , which slows down the reaction but increases the maximum conversion
5. this stages are repeated three times and then the mixture exits the reactor

ENDOTHERMIC REACTION

The case of endothermic reactions is simpler as the isoconversion curves don't intersect. ● increasing the temperature the rate of reaction increases ● increasing the temperature the maximum equilibrium conversion increases → both kinetic and thermodynamics are favored by high temperature, the only limit is the resistance of the materials.