A reaction has a rate constant of 1.23×10-4 s at 28 degrees C and 0.235 s at 79 degrees C.
A: Determine the activation barrier for the reaction.
B: What is the value of the rate constant at 15 degrees C?
-------------
I know that I need to use the Arrhenius formula. However, I am not sure which values go where. If someone could please show me, I would greatly appreciate it.
To determine the activation barrier (also called the activation energy), you 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 gas constant (8.314 J/(mol·K)),
and T is the temperature in Kelvin.
To solve for the activation energy (Ea), we need to rearrange the equation:
Ea = -ln(k / A) * (R / T)
Now let's plug in the values and solve for the activation energy at 28 degrees C (which is 28 + 273 = 301 K):
k1 = 1.23 × 10^(-4) s
T1 = 301 K
Ea = -ln(k1 / A) * (R / T1)
Next, we can solve for the activation energy at 79 degrees C (which is 79 + 273 = 352 K):
k2 = 0.235 s
T2 = 352 K
Ea = -ln(k2 / A) * (R / T2)
Now we can calculate the activation energy (Ea) using these equations.
To determine the activation barrier for the reaction, we can use the Arrhenius equation:
k1 = A * e^(-Ea/RT1)
k2 = A * e^(-Ea/RT2)
Where:
- k1 and k2 are the rate constants at temperatures T1 and T2 respectively.
- A is the pre-exponential factor.
- Ea is the activation energy.
- R is the gas constant (8.314 J/(mol·K)).
- T1 and T2 are temperatures in Kelvin.
Let's plug in the given values:
For the first scenario at 28 degrees C:
- T1 = 28 + 273 = 301 K
- k1 = 1.23 × 10^(-4) s
For the second scenario at 79 degrees C:
- T2 = 79 + 273 = 352 K
- k2 = 0.235 s
To determine the activation barrier (Ea), we can rewrite the equation as follows:
ln(k2/k1) = (-Ea/R) * (1/T2 - 1/T1)
Now let's substitute the known values:
ln(0.235 / 1.23 × 10^(-4)) = (-Ea/8.314) * (1/352 - 1/301)
Now we can solve for Ea. Rearranging the equation:
Ea = -ln(0.235 / 1.23 × 10^(-4)) / (1/8.314) * (1/352 - 1/301)
Calculating the right-hand side gives the value of Ea, which is the activation barrier for the reaction.
For part B, we can use the same Arrhenius equation:
k = A * e^(-Ea/RT)
Given that we know the rate constant at 28 degrees C (k1), we can rearrange the equation to solve for A:
A = k1 / e^(-Ea/RT1)
To find the rate constant at 15 degrees C (T), we can use the obtained value of A and the rearranged equation:
k = A * e^(-Ea/RT)
Substituting the values:
A = k1 / e^(-Ea/RT1)
k = (k1 / e^(-Ea/RT1)) * e^(-Ea/RT)
Where:
- k is the rate constant at 15 degrees C.
- Ea is the activation energy (obtained in part A).
- R is the gas constant.
- T is the temperature in Kelvin (15 + 273).
Let's plug in the known values and calculate the rate constant at 15 degrees C (T).