An alternative route to the production of propylene from propane would be
through oxydehydrogenation
At atmospheric pressure, what is the fraction of propane converted to propylene
at 400, 500, and 600°C if equilibrium is reached at each temperature? Compare
the results to those from Exercise 1. What do you think is the major impediment
to this route of olefin formation versus dehydrogenation? Assume ideal
behavior.
Temperature (0C)
400 ka :1.16* 10~26
500 Ka: 5.34 X 10~23
600 ka: 8.31 X 10~21
To determine the fraction of propane converted to propylene at each temperature, we need to use the equilibrium constant (Ka) for the given reactions. The equilibrium constant relates the concentrations of the reactants and products at equilibrium.
The equilibrium constant expression for the reaction between propane and propylene can be written as:
Ka = [propylene] / [propane]
Given the equilibrium constants for each temperature, we can calculate the fractions of propane converted to propylene.
At 400°C:
Ka = 1.16 × 10^(-26)
The fraction of propane converted to propylene can be calculated as:
Fraction of propane converted = Ka / (1 + Ka)
Substituting the value of Ka:
Fraction of propane converted at 400°C = (1.16 × 10^(-26)) / (1 + 1.16 × 10^(-26))
At 500°C:
Ka = 5.34 × 10^(-23)
Fraction of propane converted at 500°C = (5.34 × 10^(-23)) / (1 + 5.34 × 10^(-23))
At 600°C:
Ka = 8.31 × 10^(-21)
Fraction of propane converted at 600°C = (8.31 × 10^(-21)) / (1 + 8.31 × 10^(-21))
Now, let's compare the results with those from Exercise 1 to see the difference in the fraction of propane converted to propylene at each temperature.
Regarding the major impediment to the route of olefin formation through oxydehydrogenation compared to dehydrogenation, it is necessary to analyze the equilibrium constants. From the given values of the equilibrium constant (Ka) for each temperature, we can observe that the values are extremely small. This indicates that the equilibrium heavily favors the reactant side, which means that very little propane will be converted to propylene.
The major impediment to this route of olefin formation is the unfavorable equilibrium position and the very low conversion. This suggests that the oxydehydrogenation reaction does not proceed efficiently to produce propylene. The extremely small values of the equilibrium constants indicate that the forward reaction is significantly less favored compared to the reverse reaction.
To determine the fraction of propane converted to propylene at each temperature, we can use the equilibrium constant (Ka) values provided. The equilibrium constant (Ka) relates the concentrations of reactants and products at equilibrium.
The equation for the conversion of propane to propylene through oxydehydrogenation is as follows:
C3H8 ⇌ C3H6 + H2
At equilibrium, the following is true:
Ka = [C3H6] / [C3H8]
To find the fraction of propane converted to propylene, we can calculate:
Conversion = [C3H6] / ([C3H6] + [C3H8])
Let's calculate the fraction of propane converted to propylene at each temperature using the given equilibrium constant (Ka) values:
1. At 400°C:
Ka = 1.16 * 10^-26
Using the equilibrium expression, we can write:
1.16 * 10^-26 = [C3H6] / [C3H8]
Assuming ideal behavior, we can set the concentrations of C3H6 and C3H8 to be equal to 1 (as they are in moles):
1.16 * 10^-26 = 1 / 1
Therefore, the fraction of propane converted to propylene at 400°C is 1.16 * 10^-26.
2. At 500°C:
Ka = 5.34 * 10^-23
Using the equilibrium expression:
5.34 * 10^-23 = [C3H6] / [C3H8]
Again, assuming ideal behavior:
5.34 * 10^-23 = 1 / 1
Thus, the fraction of propane converted to propylene at 500°C is 5.34 * 10^-23.
3. At 600°C:
Ka = 8.31 * 10^-21
Evaluating the equilibrium expression:
8.31 * 10^-21 = [C3H6] / [C3H8]
Using ideal behavior as before:
8.31 * 10^-21 = 1 / 1
Therefore, the fraction of propane converted to propylene at 600°C is 8.31 * 10^-21.
Comparing these results to the ones calculated in Exercise 1, we can see that the fractions of propane converted to propylene using oxydehydrogenation are much smaller.
The major impediment to this route of olefin formation (oxydehydrogenation) versus dehydrogenation could be the significantly smaller equilibrium constants (Ka) for oxydehydrogenation in comparison to dehydrogenation. This implies that at equilibrium, there is a higher concentration of propane relative to propylene in the oxydehydrogenation process, leading to lower conversion rates.