The rate constant,K, for a given was determined at a number of temperatures. A graph of log K versus the reciprocal of the kelvin temperature yielded a straight line with a slope of -250 K. What is the activation energy of this reaction in Kj/mol?
For this question I got 2.08kj/mol
I don't have my calculator handy but that looks right to me.
250 * about 8 = about 2000 J/mol
To find the activation energy for a reaction given the slope of a graph, you can utilize the Arrhenius equation:
k = A * e^(-Ea/RT)
In this equation, k represents the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
First, let's analyze the given information. We have a graph of log k versus 1/T (reciprocal of the Kelvin temperature) that yields a straight line with a slope of -250 K. The slope of the line represents -Ea/R.
So, we have:
-250 K = -Ea/R
Multiplying both sides by R:
-250 K * R = -Ea
We know that R is the gas constant with a value of 8.314 J/(mol·K), but we need to convert the activation energy to KJ/mol. Dividing both sides by 1000:
(-250 K * R) / 1000 = -Ea / 1000
-250 K * 8.314 J/(mol·K) = -Ea / 1000
-2078.5 J/(mol·K) = -Ea / 1000
Now, let's convert from J to KJ:
-2078.5 J/(mol·K) = -2.0785 KJ/(mol·K)
Thus, the activation energy is approximately 2.08 KJ/mol.