What is the activation energy for a reaction whose rate is tripled by a temperature increase from 28 to 53 C? I understand that I have to use the arrhenius equation, but how do I set it up for a problem like this?
Start by writing the equation on a sheet of paper. Then write each of the variables and substitute into the equation. You will have the one unknown. Solve for that. Since you don't have rates quoted, you may EITHER use k1 for one rate constant and 3k1 for the other OR you may just make up a number for rate constant k1, then multiply that number for k2.
but I don't have the K or the Ea. Do you mind posting a sample?
ln(k2/k1) = (Ea/R)(1/T1)(1/T2)
You did not read my response very well. Make up a value for k1 if you wish, multiply that by 3 for k2, you have T1 and T2 which leaves ONLY Ea as the unknown. Don't forget to change T1 and T2 to Kelvin.
I got 896.3 KJ. Does this seem right?
actually, the right answer was 35.8 KJ
how did u get that
To set up the Arrhenius equation for this problem and determine the activation energy, we can use the equation:
k2/k1 = exp[(Ea/R) * (1/T1 - 1/T2)]
Where:
- k1 is the rate constant at temperature T1 (in Kelvin)
- k2 is the rate constant at temperature T2 (in Kelvin)
- Ea is the activation energy (in joules)
- R is the gas constant (8.314 J/(mol*K))
In this case, the rate of reaction is tripled by increasing the temperature from 28 to 53°C. Here's how you can set up the equation:
Step 1: Convert the temperatures from Celsius to Kelvin
T1 = 28°C + 273.15 = 301.15 K
T2 = 53°C + 273.15 = 326.15 K
Step 2: Take the natural logarithm to isolate the activation energy (Ea)
ln(k2/k1) = (Ea/R) * (1/T1 - 1/T2)
Step 3: Rearrange the equation to solve for Ea
Ea = ln(k2/k1) * (R / (1/T1 - 1/T2))
Step 4: Plug in the values
In this case, you need to have the rate constants (k1 and k2) for the reaction at the given temperatures (T1 and T2). If these values are not given, you won't be able to find the specific activation energy.
Once you have all the values, substitute them into the equation and calculate Ea. Remember to use consistent units throughout the calculation (usually Kelvin for temperature).