1.) At what temperature will the rate constant for the reaction H2 + I2 to 2HI have the value 5.2×10−3 M^{-1}s^{-1}? (Assume k=5.4 *10^{-4}M}{-1)s}^{-1} at 599 K and k=2.8 * 10^{-2}M}^{-1}s}^{-1} at 683 K.)
2.)For the first-order reaction
N2O5 to 2NO2 + 1/2O2
t_{1/2}=22.5 h at 20 C and 1.5 h at 40 C.
What's the activation energy
I think 1 has been done later.
#2 use the Arrhenius equation.
To answer these questions, we can use the Arrhenius equation, which relates the rate constant (k) of a reaction with the temperature (T) and the activation energy (Ea):
k = A * e^(-Ea / (R * T))
where:
- k is the rate constant of the reaction,
- A is the pre-exponential factor (also known as the frequency factor or attempt frequency),
- Ea is the activation energy,
- R is the ideal gas constant (8.314 J/(mol·K)), and
- T is the temperature in Kelvin (K).
To find the activation energy (Ea) in the first question, let's rearrange the equation as follows:
Ea = -(ln(k / A) * (R * T))
1.) At what temperature will the rate constant for the reaction H2 + I2 to 2HI have the value 5.2×10^(-3) M^(-1)s^(-1)?
Given data:
1st temperature (T1) = 599 K, k1 = 5.4×10^(-4) M^(-1)s^(-1)
2nd temperature (T2) = 683 K, k2 = 2.8×10^(-2) M^(-1)s^(-1)
To find T at a given k, we can rearrange the equation:
T = -(ln(k / A) * (R / Ea))
Let's calculate the activation energy (Ea) at the two given temperatures:
Ea = -(ln(k1 / A) * (R / T1))
Ea = -(ln(5.4×10^(-4) / A) * (8.314 / 599))
Ea = -(ln(k2 / A) * (R / T2))
Ea = -(ln(2.8×10^(-2) / A) * (8.314 / 683))
Now, we have two equations with two unknowns, Ea and A. Since the frequency factor (A) is the same for both equations, we can eliminate it by dividing the two equations:
(ln(k1 / A) * (8.314 / 599)) / (ln(k2 / A) * (8.314 / 683)) = 1
Simplifying,
(ln(k1 / A) * (8.314 / 599)) = (ln(k2 / A) * (8.314 / 683))
We can rearrange this equation to solve for A:
A = (k1 * e^(-Ea / (R * T1))) / e^(-Ea / (R * T2))
Now, we can substitute the known values to determine A:
A = (5.4×10^(-4) * e^(-Ea / (8.314 * 599))) / e^(-Ea / (8.314 * 683))
After finding the value of A, substitute it into one of the previous equations to find the value of Ea:
Ea = -(ln(k1 / A) * (8.314 / 599))
2.) For the first-order reaction N2O5 to 2NO2 + 1/2O2, we are given the half-life (t_1/2) at two different temperatures (T1 and T2).
The half-life of a first-order reaction can be related to the rate constant (k) and the temperature (T) using the following equation:
t_1/2 = (ln(2)) / k
The Arrhenius equation can be rearranged to solve for k:
k = A * e^(-Ea / (R * T))
Since we need to find the activation energy (Ea), we can manipulate the equation as follows:
Ea = -(ln(k / A) * (R * T))
Let's use the given half-life values to first solve for the rate constant (k) at each temperature.
t_1/2 = (ln(2)) / k
k1 = (ln(2)) / t_1/2 at T1
k2 = (ln(2)) / t_1/2 at T2
Then, use the values of k1 and k2, along with the temperatures T1 and T2, to solve for Ea. You can rearrange the equation:
Ea = -(ln(k / A) * (R / T))
to solve for Ea:
Ea = -(ln(k1 / A) * (R / T1))
Ea = -(ln(k2 / A) * (R / T2))
Now, you can solve the two equations simultaneously to find the activation energy Ea.