The presence of a catalyst provides a reaction pathway in which the activation energy of a certain reaction is reduced from 188 kJ/mol to 86 kJ/mol which increases the rate of the reaction.
How many times faster is the reaction with the catalyst at 307 K, all other factors being equal?
POV: you get a response almost a decade later, you've probably already graduated, have a family of your own and moved on from school. Anyways the answers are A, C, C, D.
To determine how many times faster the reaction is with the catalyst, we need to compare the rates of the reaction with and without the catalyst.
The rate of a reaction can be expressed using the Arrhenius equation:
Rate = A * e^(-Ea/RT)
where:
- Rate is the reaction rate
- A is the pre-exponential factor
- Ea is the activation energy
- R is the gas constant (8.314 J/(mol*K))
- T is the temperature in Kelvin
Let's assume that the pre-exponential factor and all other factors are the same for both reactions, as stated in the question.
Without the catalyst:
Rate_1 = A * e^(-Ea1/RT)
With the catalyst:
Rate_2 = A * e^(-Ea2/RT)
We know that Ea1 = 188 kJ/mol and Ea2 = 86 kJ/mol. The temperature is given as 307 K.
Now, we can divide the rates to determine how many times faster the reaction is with the catalyst:
Rate_2 / Rate_1 = (A * e^(-Ea2/RT)) / (A * e^(-Ea1/RT))
The pre-exponential factor and other factors cancel out, leaving us with:
Rate_2 / Rate_1 = e^(-(Ea2 - Ea1)/(RT))
Substituting the values, we get:
Rate_2 / Rate_1 = e^(-((86 - 188) * 1000)/(8.314 * 307))
Calculating the expression inside the exponential and the exponential itself, we find:
Rate_2 / Rate_1 = e^(102000/(2550.442))
Finally, we can calculate the result using a calculator or a math library:
Rate_2 / Rate_1 ≈ 24.821
Therefore, the reaction is approximately 24.821 times faster with the catalyst at 307 K.