The reaction rate triples when the temperature increases from 15 to 35 degree C. What is the activation energy?
I believe you use the Arrhenius equation. Make the k values a ratio of 3.
From Arhenius Equation
ln(k1/k2)= Ea/R((T1-T2)/(T1T2))
The ratio of rate constant is 1:3;
R= 8.314JK.mole
T1=288K
T2=308K
So;
ln(1/3)=Ea/8.314((288-308)/(288x308))
Ea=4.05x10⁴J or 4.05x10¹Kj
To determine the activation energy, we can use the Arrhenius equation, which relates the reaction rate constant (k) to the temperature (T) and activation energy (Ea):
k = A * exp(-Ea / (R * T))
Where:
- k is the reaction rate constant
- A is the pre-exponential factor or frequency factor (constant)
- Ea is the activation energy
- R is the gas constant (8.314 J/(mol*K))
- T is the absolute temperature in Kelvin
To begin, let's convert the given temperatures into Kelvin:
- T1 = 15°C + 273.15 = 288.15 K
- T2 = 35°C + 273.15 = 308.15 K
We are told that the reaction rate triples, which means k2/k1 = 3.
Now, we can set up the ratio of the rate constants and solve for the activation energy (Ea):
3 = (A * exp(-Ea / (R * T2))) / (A * exp(-Ea / (R * T1)))
Canceling out the A and rearranging the equation, we get:
3 = exp((-Ea / (R * T2)) + (Ea / (R * T1)))
Taking the natural logarithm (ln) of both sides to eliminate the exponential term:
ln(3) = (-Ea / (R * T2)) + (Ea / (R * T1))
Rearranging the equation to solve for Ea:
ln(3) * R = Ea * ((1 / (R * T1)) - (1 / (R * T2)))
Simplifying further:
Ea = (ln(3) * R) / ((1 / (R * T1)) - (1 / (R * T2)))
Plugging in the values:
Ea = (ln(3) * 8.314 J/(mol*K)) / ((1 / (8.314 J/(mol*K) * 288.15 K)) - (1 / (8.314 J/(mol*K) * 308.15 K)))
Ea ≈ 39.7 kJ/mol
Therefore, the activation energy is approximately 39.7 kJ/mol.