A certain reaction has an activation energy of 74.38 kJ/mol. At what Kelvin temperature will the reaction proceed 7.00 times faster than it did at 313 K?
Use the Arrhenius equation. Let k1 = k1 and 7k1 = k2.
To find the Kelvin temperature at which the reaction will proceed 7.00 times faster, we need to use the Arrhenius equation, which relates the rate constant of a reaction (k) to the activation energy (Ea) and the temperature in Kelvin (T).
The Arrhenius equation is given by:
k = A * exp(-Ea / (R * T))
Where:
k is the rate constant
A is the pre-exponential factor (or frequency factor)
Ea is the activation energy
R is the gas constant (8.314 J/(mol*K))
T is the temperature in Kelvin
We know that at a higher temperature, the reaction will proceed faster. And we are given that the reaction will proceed 7.00 times faster at a certain temperature, compared to 313 K. Let's call this temperature T2.
To find T2, we need to solve the following equation:
k2 = 7 * k1
Where k2 is the rate constant at T2 and k1 is the rate constant at 313 K.
Now, let's substitute the Arrhenius equation into the equation for k2:
A * exp(-Ea2 / (R * T2)) = 7 * (A * exp(-Ea1 / (R * 313)))
We can simplify and cancel out the A factor:
exp(-Ea2 / (R * T2)) = 7 * exp(-Ea1 / (R * 313))
Taking the logarithm of both sides:
-Ea2 / (R * T2) = ln(7) - Ea1 / (R * 313)
Rearranging the equation:
Ea2 = -R * T2 * (ln(7) - Ea1 / (R * 313))
Simplifying:
T2 = (-Ea2 / (R * (ln(7) - Ea1 / (R * 313))))
Now we can substitute the given values to find the temperature T2:
Ea1 = 74.38 kJ/mol = 74.38 * 1000 J/mol
Ea2 = Ea1
R = 8.314 J/(mol*K)
T1 = 313 K
T2 = (-Ea2 / (R * (ln(7) - Ea1 / (R * 313))))
= (-74.38 * 1000 / (8.314 * (ln(7) - 74.38 * 1000 / (8.314 * 313))))
Calculating this value will give you the Kelvin temperature at which the reaction will proceed 7.00 times faster.