At what temperature will the rate constant for the reaction have the value 6.5×10−3 M/s?
(Assume k = 5.4 * 10^-4 M/s-1 at 599K and
k = 2.8 * 10^-2 M/s at 683K .)
You need to use the Arrhenius equation twice. The first time use it with the two k values given with the two T values and solve for Ea (activation energy), then use it the second time using Ea from the first calcn and either one of the K and T values to solve for T value for the desired k value in the problem.
Hey I tried that and still cant get the right answer. The answer that I got is 555.6 or 556 K and its not correct. What answer did you get?
I didn't do it. If you will post your work I will find the error.
To find the temperature at which the rate constant has a specific value, we can use the Arrhenius equation. The Arrhenius equation relates the rate constant (k) to the temperature (T) and the activation energy (Ea) of a reaction:
k = A * exp(-Ea/RT)
Where:
- k is the rate constant
- A is the pre-exponential factor (frequency factor)
- Ea is the activation energy
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
In this case, we are given two different rate constants at two different temperatures. So, we can set up two equations using the Arrhenius equation and solve for T using algebraic manipulation.
Let's denote the given rate constants as k1 = 5.4 * 10^-4 M/s-1 at T1 = 599K and k2 = 2.8 * 10^-2 M/s at T2 = 683K. We need to find the temperature at which k = 6.5×10−3 M/s.
Equation 1: k1 = A * exp(-Ea/RT1)
Equation 2: k2 = A * exp(-Ea/RT2)
Dividing equation 2 by equation 1, we get:
k2/k1 = (A * exp(-Ea/RT2)) / (A * exp(-Ea/RT1))
Simplifying:
k2/k1 = exp(-Ea/RT2) / exp(-Ea/RT1)
Since exp(a) / exp(b) = exp(a - b), we have:
k2/k1 = exp(Ea/RT1 - Ea/RT2)
Taking the natural logarithm of both sides:
ln(k2/k1) = Ea/RT1 - Ea/RT2
Now, we can solve for Ea/ R by rearranging the equation:
Ea/R = ln(k2/k1) / (1/T1 - 1/T2)
Finally, we can rearrange the Arrhenius equation to find the temperature (T) for a specific rate constant (k):
T = Ea / (R * (ln(k) - ln(A)))
Now, substitute the given values:
- k = 6.5×10−3 M/s
- ln(k) = ln(6.5×10−3 M/s)
- ln(A) = ln of the pre-exponential factor (which is not given)
- Ea = Ea/R calculated earlier
Plug in these values into the equation, and you will get the temperature at which the rate constant has the value 6.5×10−3 M/s.