For the gas phase decomposition of chloroethane,
CH3CH2Cl C2H4 + HCl
the rate constant at 723 K is 1.75E-4 /s and the rate constant at 769 K is 2.19E-3 /s.
The activation energy for the gas phase decomposition of chloroethane is
USe the combined arrehiuns equation and solve for Ea:
ln(k1/k2 )=Ea/R(1/T2−1/T1)
Where
k1=1.75 x 10^-4
T1=723K
k2=2.19 x 10^-3
T2=769K
R= 8.314 J/mol*K
And
Ea=?
ln(1.75 x 10^-4/2.19 x 10^-3 )=Ea/8.314 J/mol*K(1/769K−1/723K)
-2.5269=Ea/8.314 J/mol*K(-8.2736 x 10^-5)
-2.5269/-8.2736 x 10^-5=Ea/8.314 J/mol*K
3.054 x 10^4=Ea/8.314 J/mol*K
Ea=3.054 x 10^4*8.314 J/mol*K
Ea=2.54 x 10^5 J/mol
To find the activation energy for the gas phase decomposition of chloroethane, we can use the Arrhenius equation:
k = A * e^(-Ea / RT)
where:
- k is the rate constant
- A is the pre-exponential factor
- Ea is the activation energy
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
We are given two sets of rate constants and temperatures, so we can set up two equations:
k1 = A * e^(-Ea / RT1)
k2 = A * e^(-Ea / RT2)
where:
- k1 is the rate constant at the first temperature (723 K)
- k2 is the rate constant at the second temperature (769 K)
- T1 and T2 are the first and second temperatures respectively
We can divide these two equations to eliminate the pre-exponential factor, A:
k1 / k2 = (A * e^(-Ea / RT1)) / (A * e^(-Ea / RT2))
k1 / k2 = e^((-Ea / RT1) + (Ea / RT2))
k1 / k2 = e^((Ea / R) * (1/RT2 - 1/RT1))
Now we have an equation with only the activation energy and temperatures. We can rearrange this equation to solve for Ea:
ln(k1 / k2) = (Ea / R) * (1/RT2 - 1/RT1)
Let's plug in the given values:
k1 = 1.75E-4 /s
k2 = 2.19E-3 /s
T1 = 723 K
T2 = 769 K
R = 8.314 J/(mol·K)
Now we can calculate the activation energy (Ea):
ln(1.75E-4 / 2.19E-3) = (Ea / 8.314) * (1/769 - 1/723)
Simplifying the equation:
-6.0636 = (Ea / 8.314) * (0.001302 - 0.001381)
-6.0636 = (Ea / 8.314) * (-7.9E-5)
Now we can solve for Ea:
Ea = -6.0636 / ((-7.9E-5) * (8.314))
Ea ≈ 73,954 J/mol
Therefore, the activation energy for the gas phase decomposition of chloroethane is approximately 73,954 J/mol.
To calculate the activation energy for the gas phase decomposition of chloroethane, you can use the Arrhenius equation. The Arrhenius equation relates the rate constant (k) of a reaction to the temperature (T) and the activation energy (Ea) of the reaction:
k = A * exp(-Ea / RT)
In this equation, k is the rate constant, T is the temperature in Kelvin, R is the gas constant (8.314 J/(mol*K)), Ea is the activation energy in Joules (J), and A is the pre-exponential factor or frequency factor.
To solve for the activation energy (Ea), you will need to rearrange the equation as follows:
Ea = -R * ln(k / A) * T
Now, let's calculate the activation energy for chloroethane at 723 K and 769 K using the given rate constants:
1. For the rate constant at 723 K (k1 = 1.75E-4 /s):
Substitute the values into the equation:
Ea1 = -8.314 J/(mol*K) * ln(1.75E-4 /s / A) * 723 K
2. For the rate constant at 769 K (k2 = 2.19E-3 /s):
Substitute the values into the equation:
Ea2 = -8.314 J/(mol*K) * ln(2.19E-3 /s / A) * 769 K
Since the pre-exponential factor (A) remains constant, we can take the ratio of the two equations to eliminate A:
Ea1 / Ea2 = (ln(k1 / A) * T2) / (ln(k2 / A) * T1)
Simplifying the equation:
Ea1 / Ea2 = (ln(k1) - ln(A)) * T2 / (ln(k2) - ln(A)) * T1
Since the pre-exponential factor (A) is typically very small, it can be assumed negligible. Therefore, we can simplify the equation further:
Ea1 / Ea2 ≈ ln(k1) * T2 / ln(k2) * T1
Now, substitute the given values into the equation:
Ea1 / Ea2 ≈ ln(1.75E-4 /s) * 769 K / ln(2.19E-3 /s) * 723 K
Evaluate the right-hand side of the equation using natural logarithms (ln):
Ea1 / Ea2 ≈ -11.967 * 769 K / -6.124 * 723 K
Finally, solve for the activation energy ratio:
Ea1 / Ea2 ≈ 19.36
Therefore, the activation energy for the gas phase decomposition of chloroethane is approximately 19.36 times larger at 723 K compared to 769 K.