The rate constant nearly doubles for some reaction when there is an increase in temperaure from 17 degree celsius to 32 degree celsius. Calculate the activation energy.
Use the Arrhenius equation. Set k1 at any number and k2 will be double that number OR set k2 = 2k1.
I am still confused could you work the problem out.
To calculate the activation energy, we can make use of the Arrhenius equation. The Arrhenius equation is given by:
k = Ae^(-Ea/RT)
Where:
k = rate constant
A = pre-exponential factor (also known as the frequency factor)
Ea = activation energy
R = ideal gas constant (8.314 J/(mol*K))
T = temperature in Kelvin
In this case, we are given the rate constant at two different temperatures, so we can set up two equations using the Arrhenius equation and then solve for the activation energy.
Let's use T1 = 17°C = 17 + 273.15 = 290.15 K and T2 = 32°C = 32 + 273.15 = 305.15 K.
We are given that the rate constant doubles when the temperature increases from T1 to T2. This means that k2 = 2 * k1.
Now, we can set up two equations using the Arrhenius equation:
k1 = A * e^(-Ea / (R * T1))
k2 = A * e^(-Ea / (R * T2))
Since we know that k2 = 2 * k1, we can substitute this into the second equation:
2 * k1 = A * e^(-Ea / (R * T2))
Now we can solve for the activation energy (Ea):
k1 = A * e^(-Ea / (R * T1))
Rearranging the equation for Ea, we have:
Ea = -R * T1 * ln(k1 / A)
Substituting the given values:
Ea = -8.314 J/(mol*K) * 290.15 K * ln(k1 / A)
Now, you can substitute the actual values of k1 and A into this equation, and then evaluate the natural logarithm.