a chemist runs a chemical reaction at 25 celsius and decides that it proceeds far too repedily. As a result he decides that the reaction rate must be decreased by a factor of 4. At what teperature should the chemist run the reaction to achieve this goal
see other post
To decrease the reaction rate by a factor of 4, we can use the Arrhenius equation, which relates the rate constant (k) of a reaction to the temperature (T). The Arrhenius equation is:
k = A * e^(-Ea/RT)
Where:
- k is the rate constant
- A is the pre-exponential factor (a constant unique to each reaction)
- e is the base of natural logarithm (approximately equal to 2.71828)
- Ea is the activation energy of the reaction
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
Since we need to decrease the reaction rate by a factor of 4, we can rearrange the equation as follows:
k' = (1/4) * k
Now we can substitute this into the Arrhenius equation:
(1/4) * k = A * e^(-Ea/RT)
Next, we need to isolate the exponential term on one side:
e^(-Ea/RT) = (4/A) * k
To get the temperature at which the reaction proceeds at the desired rate, we need to solve for T.
First, take the natural logarithm (ln) of both sides of the equation:
ln(e^(-Ea/RT)) = ln((4/A) * k)
The natural logarithm of e^(-Ea/RT) cancels out:
-Ea/RT = ln((4/A) * k)
Rearrange the equation for T:
T = -Ea / (R * ln((4/A) * k))
To find the temperature (T) at which the reaction should be run, we need the activation energy (Ea), the pre-exponential factor (A), and the rate constant (k).