Propylene can be produced from the dehydrogenation of propane over a catalyst.
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? Assume ideal behavior.
(Temperature(C)
400 ka:0.000521
500 ka:0.0104
600 ka:0.104
At 600.
Take 100 mols as an arbitrary number.
.......CH3CH3CH3 ==> CH3CH=CH2 + H2
I.......100............0..........0
C........-x............x..........x
E......100-x...........x..........x
Ka = 0.104 = (x)(x)/(100-x)
Solve for x for mols = x, then
(x/100) = fraction converted.
I estimate about 14% but check me out on that.
To determine the fraction of propane converted to propylene at different temperatures, we need to use the equilibrium constant (K) and the given equilibrium constants at each temperature (ka). The equilibrium constant is defined as the ratio of the product concentrations to the reactant concentrations, raised to the power of their stoichiometric coefficients.
The general equation for the dehydrogenation of propane, assuming ideal behavior, can be represented as:
2C3H8 ⇌ C3H6 + H2
Let's calculate the fraction of propane converted to propylene at 400°C, 500°C, and 600°C using the given equilibrium constants:
1. At 400°C (ka = 0.000521):
K = (C3H6 * H2) / (C3H8^2)
Let's assume x represents the fractional conversion of propane to propylene.
At equilibrium:
(C3H6 * H2) / (C3H8^2) = ka
(C3H6 * H2) / (x^2) = ka
C3H6 * H2 = ka * (x^2)
Since we assume ideal behavior, we can write:
(C3H8_initial - C3H8_converted) ≈ C3H8_initial
(x) ≈ 1 - (C3H8_converted / C3H8_initial)
Now, we can substitute the values and solve for x:
(x) ≈ 1 - (C3H8_converted / C3H8_initial)
(x) ≈ 1 - (C3H6 * H2) / (ka * C3H8_initial^2)
(x) ≈ 1 - (ka * (x^2)) / (ka * C3H8_initial^2)
(x) ≈ 1 - (x^2) / C3H8_initial
Substituting C3H8_initial = 1 (since it is the fraction of propane), we get:
(x) ≈ 1 - (x^2) / 1
(x) ≈ 1 - (x^2)
Solving the quadratic equation:
x^2 + x - 1 = 0
Using the quadratic formula, we find:
x = (-1 ± √5) / 2
Since we are looking for the fraction of propane converted, we discard the negative root and take the positive root:
x ≈ (√5 - 1) / 2
Therefore, at 400°C, the fraction of propane converted to propylene is approximately (√5 - 1) / 2.
You can follow the same steps for the other temperatures (500°C and 600°C) using the respective ka values:
2. At 500°C (ka = 0.0104):
Using the same calculations, the fraction of propane converted to propylene is approximately (√41 - 1) / 8.
3. At 600°C (ka = 0.104):
The fraction of propane converted to propylene is approximately (√413 - 1) / 18.
To determine the fraction of propane converted to propylene at equilibrium, we will use the equilibrium constant (Ka) values provided for each temperature. The equilibrium constant is a measure of how far a reaction goes to reach equilibrium.
The equilibrium constant expression for the dehydrogenation reaction of propane to propylene can be written as:
Ka = ([C3H6]/[C3H8])
Where [C3H6] represents the concentration of propylene and [C3H8] represents the concentration of propane.
To find the fraction of propane converted to propylene, we need to calculate the concentration of propylene relative to the initial concentration of propane.
Let's assume we start with 1 mole of propane.
At 400°C (T = 400):
Ka = 0.000521
To find the fraction of propane converted to propylene, we can use the equilibrium constant expression:
Ka = ([C3H6]/[C3H8])
Since we start with 1 mole of propane and the fraction of converted propane is x:
Ka = ([x]/[1-x])
We can rearrange the equation:
0.000521 = (x)/(1-x)
Solving for x, we find:
x = 0.000521/(1 + 0.000521) = 0.00052048
So, at 400°C, the fraction of propane converted to propylene is approximately 0.00052048.
Similarly, we can repeat the above calculations for the other two temperatures.
At 500°C (T = 500):
Ka = 0.0104
x = 0.0104/(1 + 0.0104) = 0.01029803
At 600°C (T = 600):
Ka = 0.104
x = 0.104/(1 + 0.104) = 0.09424603
Therefore, at 400°C, the fraction of propane converted to propylene is approximately 0.00052048, at 500°C it is approximately 0.01029803, and at 600°C it is approximately 0.09424603.