Consider the industrial production of cyclohexane, C6H12, by the gas-phase hydro
ID: 701389 • Letter: C
Question
Consider the industrial production of cyclohexane, C6H12, by the gas-phase hydrogenation of benzene, C6H6. Assume that this process is carried out by two reactors in series as shown below. The first reactor is at 340°C and 5 bar, while the second reactor is at 265°C and 5 bar. The feed ratio of hydrogen gas to benzene is 101; there is no cyclohexane in the feed. All species are in the gas phase. You may assume ideal gas behavior and that Dhrxn does not change with temperature.
(a) What is the equilibrium composition at the exit of the second reactor?
(b) What is the purpose of the first reactor; that is, why do we use two reactors instead of just one?
(c) Would we get more product if we used a pressure of 1 bar instead of 5 bar. Explain.
(d) Would you recommend diluting the feed with an inert to increase the yield of C6H12? Explain.
C6H12 is obtained in product.
10 moles/sec H2(g) 1 moles/sec C6H6(g) T= 340°C P= 5 bar Reactor1 T= 265°C P=5bar Reactor2Explanation / Answer
a) At the outlet of 1st reactor, In equilibrium, Let's consider
Molar flow rate of C612 = a
Molar flowrate of C6H6 = b
Molar flowrate fo H2 = c
Reaction is, C6H6 + 3H2 = C6H12
Therefore, one mole fo C6H6 gives 1 mole of C6H12 and consumes 3 moles of H2
Since, Hydrogen inlet molar flowrate = 10 mol/s
Benzene inlet molar flow rate = 1 mol/s
Therefore,
1-b = a
10-3*(1-b) = c
Therfore, c=7+3b
Total flowrate at outlet of 1st reactor = a+b+c = 1-b+b+7+3b = 8+3b
Mole fraction of C6H12 in 1st reactor at equilibrium = (1-b)/(8+3b)
Mole fraction of C6H6 in 1st reactor at equilibrium = (b)/(8+3b)
Mole fraction of H2 in 1st reactor at equilibrium = (7+3b)/(8+3b)
Since, Total Pressure of reactor 1 = 5 bar
Partial Pressure of C6H12 in 1st reactor at equilibrium = (1-b)/(8+3b)*5
Partial Pressure of C6H6 in 1st reactor at equilibrium = (b)/(8+3b)*5
Partial Pressure of H2 in 1st reactor at equilibrium = (7+3b)/(8+3b)*5
Therefore, Equilibrium constant at reactor 1, K1 = ((1-b)/(8+3b)*5)/((b)/(8+3b)*5*((7+3b)/(8+3b)*5)^3)
Similarly, At the outlet of 2nd reactor, In equilibrium, Let's consider
Molar flow rate of C612 = d
Molar flowrate of C6H6 = e
Molar flowrate fo H2 = f
Since, Hydrogen inlet molar flowrate = (7+3b) mol/s
Benzene inlet molar flow rate = b mol/s
Therefore,
d=b-e+1-b = 1-e
7+3b-3*(b-e) = f
Therfore, f=7+3e
Total flowrate at outlet of 2nd reactor = d+e+f = 8+3e
Mole fraction of C6H12 in 2nd reactor at equilibrium = (1-e)/(8+3e)
Mole fraction of C6H6 in 2nd reactor at equilibrium = (e)/(8+3e)
Mole fraction of H2 in 2nd reactor at equilibrium = (7+3e)/(8+3e)
Since, Total Pressure of reactor 2 = 5 bar
Partial Pressure of C6H12 in 2nd reactor at equilibrium = (1-e)/(8+3e)*5
Partial Pressure of C6H6 in 2nd reactor at equilibrium = (e)/(8+3e)*5
Partial Pressure of H2 in 2ndrea ctor at equilibrium = (7+3e)/(8+3e)*5
Therefore, Equilibrium constant at reactor 2, K2 = ((1-e)/(8+3e)*5)/((e)/(8+3e)*5*((7+3e)/(8+3e)*5)^3)
Now, Since no Thermodynamic data to calculate equlibrium constant is given, assuming C6H6 to be totally consumed after reactor 2. Therfore, e=0
Therefore, Outlet concentration at 2nd reactor of benzene = 0 mole/hr
Outlet concentration at 2nd reactor of C6H12 = 1mole/hr
Outlet concentration at 2nd reactor of H2 = 7 mole/hr
b) Since, reaction is in equilibrium, Whole conversion of Benzene won't occur. To increase the conversion,
another reactor is kept in series at different temperature to increase the Benzene conversion to C6H12
From equilibrium constant equation, This can be said that molar flowrate of benzene decreases by decreasing equilibrium constant.
Equilibrium constant decreases by decreasing temperature
Therefore, 2nd reactor at lower temperature to reduce benzene molar flowrate which equivalently increases desired product C6H12 molar flowrate
c) At Pressure P of reactor 2 Reaction equilibrium constant, K2 = ((1-e)/(8+3e)*P)/((e)/(8+3e)*P*((7+3e)/(8+3e)*5)^P) = 1/(P^3)*{.....}
Therefore reaction equilbrium constant is inversly proportional to Pressure
Hence, decreasing Pressure will increase equilibrium constant
More the equilibrium constant, more the reactant conversion and product formation
Therefore, we will get more product by reducing Pressure to 1 bar from 5 bar
d) Let molar flowrate of Inert = g mole/hr
Then equilibrium constant K2 = ((1-e)/(8+3e+g)*P)/((e)/(8+3e+g)*P*((7+3e)/(8+3e+g)*5)^P) = (8+3e+g)^3*{...}
Therefore, equilibrium constant increases with inert addition.
Therefore, it is recommended to use inert for better conversion to C6H12