The rate constant is 9.50x10^-5 L/mol*s at 195 degrees Celsius and 1.30x10^-3 L/mol*s at 258 degrees Celsius. What is the activation energy of the reaction?
I'm not sure what I'm doing wrong here.
I found the natural log (ln) of: 9.50x10^-5 = -9.26
1.30x10^-3 = -6.65
and took the inverse of the temperatures:
195 = 5.13x10^-3
258 = 3.88x10^-3.
Then:
(-6.65-(-9.26) / ((3.88x10^-3)-(5.13x10^-3))) = -2088K
Ea (activation energy) is equal to slope so I did Ea = -(slope)(R = 8.314) which gave me 17kJ/mol. But I know the answer is SUPPOSED to be 85.8kJ/mol.
So what did I do wrong here? I followed the instructions in my textbook to the T and yet it came out horrible wrong.
It seems like you made a mistake in your calculations. Let's go over the correct steps to find the activation energy.
First, we need to calculate the slope of the ln(rate constant) versus (1/temperature) graph. To do this, we'll use the equation:
ln(k2/k1) = -Ea/R * (1/T2 - 1/T1)
where k1 and k2 are the rate constants at different temperatures, Ea is the activation energy, R is the gas constant (8.314 J/mol*K), T1 and T2 are the corresponding temperatures.
Using the values given, the equation becomes:
ln(1.30x10^-3 / 9.50x10^-5) = -(Ea/8.314) * (1/258 - 1/195)
To solve for Ea, rearrange the equation:
-(Ea/8.314) = ln(1.30x10^-3 / 9.50x10^-5) / (1/258 - 1/195)
Now, plug in the values:
-(Ea/8.314) = ln(1.30x10^-3 / 9.50x10^-5) / (0.003876 - 0.005128)
Calculate the right side of the equation:
-(Ea/8.314) = ln(1.369) / (-0.001252)
Now, solve for Ea (activation energy):
Ea/8.314 = ln(1.369) / 0.001252
Ea = (8.314 * ln(1.369)) / 0.001252
Ea ≈ 85.8 kJ/mol
Therefore, the correct activation energy for the reaction is approximately 85.8 kJ/mol.