A particular reaction has an activation energy of 51.6 kJ/mol-rxn. How much faster is the reaction at 50 oC versus 25 oC?
a. 2 times b. 5 times c. 3 times d. half as fast d. 20% as fast
I think it would be A??
The rule of thumb is 2x faster for each 10 C increase. But with the activation energy you can use the Arrhenius equation and calculate it.
So based of the rule of thumb, its going to be 5 times faster. Also how can I use Arrhenius equation without the rate constant (k)?
To determine how much faster a reaction is at a higher temperature, we need to use the Arrhenius equation. The Arrhenius equation relates the rate constant of a reaction to the temperature and the activation energy.
The Arrhenius equation is given by: k = Ae^(-Ea/RT), where k is the rate constant, Ea is the activation energy, R is the gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin.
First, let's convert the activation energy from kJ/mol to J/mol:
51.6 kJ/mol = 51,600 J/mol
Now, we can compare the rate constant at two different temperatures.
For 25 oC:
Convert to Kelvin: T1 = 25 + 273 = 298 K
For 50 oC:
Convert to Kelvin: T2 = 50 + 273 = 323 K
Let's calculate the rate constant at each temperature using the Arrhenius equation:
k1 = A*e^(-Ea/RT1)
k2 = A*e^(-Ea/RT2)
Now, we want to find the ratio of the rate constants, k2/k1. The exponential factors cancel out, leaving us with:
(k2/k1) = e^((-Ea/R)(1/T2 - 1/T1))
Calculating these values, we get:
(k2/k1) = e^((-51600 J/mol)/(8.314 J/(mol·K)) * (1/323 K - 1/298 K))
Using a calculator, we find that (k2/k1) is approximately 5.4.
Since we want to find how much faster the reaction is at 50 oC compared to 25 oC, we can say that the reaction at 50 oC is approximately 5.4 times faster than at 25 oC.
Therefore, the correct answer is b. 5 times.