The production of ethylbenzene, a very popular industrial chemical, is carried out reacting benzene with ethylene in liquid phas
e. This reaction takes place in a series of reactors that involve multiple side reactions and intermediates. Ethylene, being the limiting reactant, is used up first and hence a considerable amount of benzene remains unreacted. From one of the reactors in the series, the exit stream is a mixture of this unreacted benzene (1), an intermediate – toluene (2), and the product ethyl benzene (3).
It is desirable to separate this liquid mixture before sending the components to the next series of reactors/process steps. So 100 mol/min of this mixture is flashed from 200 mm Hg and 50 °C to 100 mm Hg. If the mole fraction of benzene and toluene are 40% each when the mixture enters the flash distillation unit, determine if the mixture will flash completely, partially, or not at all. Assume ideal gas and ideal solution behavior for the vapor phase and liquid phase, respectively. If the mixture does flash partially, determine the composition and molar flow rates of the equilibrium streams exiting the reactor. Show all calculations by hand using your preferred method for solving simultaneous equations. Alternatively, you may use Solver (Excel) but this must be accompanied by a printout of a neatly formatted Excel sheet showing your equations and constraints.
It is desirable to separate this liquid mixture before sending the components to the next series of reactors/process steps. So 100 mol/min of this mixture is flashed from 200 mm Hg and 50 °C to 100 mm Hg. If the mole fraction of benzene and toluene are 40% each when the mixture enters the flash distillation unit, determine if the mixture will flash completely, partially, or not at all. Assume ideal gas and ideal solution behavior for the vapor phase and liquid phase, respectively. If the mixture does flash partially, determine the composition and molar flow rates of the equilibrium streams exiting the reactor. Show all calculations by hand using your preferred method for solving simultaneous equations. Alternatively, you may use Solver (Excel) but this must be accompanied by a printout of a neatly formatted Excel sheet showing your equations and constraints.
1 answer:
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Taking into account the definition of calorimetry, sensible heat and latent heat, the amount of heat required is 37.88 kJ.
<h3>Calorimetry</h3>Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
<h3>Sensible heat</h3>Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).
<h3>Latent heat</h3>Latent heat is defined as the energy required by a quantity of substance to change state.
When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.
- <u><em>25.60 °C to 0 °C</em></u>
First of all, you should know that the freezing point of water is 0°C. That is, at 0°C, water freezes and turns into ice.
So, you must lower the temperature from 25.60°C (in liquid state) to 0°C, in order to supply heat without changing state (sensible heat).
The amount of heat a body receives or transmits is determined by:
Q = c× m× ΔT
where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case, you know:
- c= Heat Capacity of Liquid= 4.184
- m= 185.5 g
- ΔT= Tfinal - Tinitial= 0 °C - 25.60 °C= - 25.6 °C
Replacing:
Q1= 4.184 × 185.5 g× (- 25.6 °C)
Solving:
<u><em>Q1= -19,868.98 J</em></u>
- <u><em>Change of state</em></u>
The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to
Q = m×L
where L is called the latent heat of the substance and depends on the type of phase change.
In this case, you know:
n= 185.5 grams× = 10.30 moles, where 18
is the molar mass of water, that is, the amount of mass that a substance contains in one mole.
ΔHfus= 6.01
Replacing:
Q2= 10.30 moles×6.01
Solving:
<u><em>Q2=61.903 kJ= 61,903 J</em></u>
- <u><em>0 °C to -10.70 °C</em></u>
Similar to sensible heat previously calculated, you know:
- c = Heat Capacity of Solid = 2.092
- m= 185.5 g
- ΔT= Tfinal - Tinitial= -10.70 °C - 0 °C= -10.70 °C
Replacing:
Q3= 2.092 × 185.5 g× (-10.70) °C
Solving:
<u><em>Q3= -4,152.3062 J</em></u>
<h3>Total heat required</h3>The total heat required is calculated as:
Total heat required= Q1 + Q2 +Q3
Total heat required=-19,868.98 J + 61,903 J -4,152.3062 J
<u><em>Total heat required= 37,881.7138 J= 37.8817138 kJ= 37.88 kJ</em></u>
In summary, the amount of heat required is 37.88 kJ.
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Answer:
Explanation:
Hello,
In this case, for the given reaction, the equilibrium constant turns out:
Nonetheless, we are asked for the reverse equilibrium constant that is:
Which is greater than one.
In such a way, the Gibbs free energy turns out:
Now, since the reverse equilibrium constant is greater than zero its natural logarithm is positive, therefore with the initial minus, the Gibbs free energy is less than zero, that is, negative.