Do you mind posting up an example for that please? It would help out a out: What is the activation energy for a reaction whose rate tripled when temperature rose from 28 to 53C.
You want me to do your work for you. I don't know what trouble you're having but you need to learn to do it yourself. It's a simple matter of writing the equation, which I posted, and substituting the numbers. Substituting into the equation can't be THAT hard to do. The only thing for you to do is
1. Substitute the numbers into the equation.
2. Solve for Ea, the only unknown in the equation.
I just read your responses to your first post. If you will show your work someone will be happy to help find the error.
Of course! To find the activation energy for a reaction, we can use the Arrhenius equation, which relates the reaction rate constant (k) to the temperature (T) and the activation energy (Ea):
ln(k2/k1) = (-Ea/R) * (1/T2 - 1/T1)
Where:
- k2 and k1 are the rate constants at temperatures T2 and T1, respectively.
- Ea is the activation energy.
- R is the gas constant (8.314 J/(mol·K)).
- T2 and T1 are the temperatures in Kelvin.
In this case, we can use the given information:
- The rate tripled, which means k2 is three times k1.
- T1 = 28°C = 28 + 273 = 301 K.
- T2 = 53°C = 53 + 273 = 326 K.
Now, let's substitute the given values into the equation:
ln(3) = (-Ea/8.314) * (1/326 - 1/301)
To determine the activation energy, we need to solve for Ea. Rearranging the equation:
Ea = -8.314 * ln(3) / (1/326 - 1/301)
Calculating the above expression will give us the activation energy for this reaction.