If the reaction rate increases by 2.6 when the temperature increases from 20-30 degrees Celsius, what is Ea?

Do we use 2.6 in place of (k2/k1) or multiply something by 2.6? Lost, please help.

yes, k2/k1 is 2.6

Ea=R*t1*T2*ln(k2/k1) / (T1-t2)

To determine the value of the activation energy (Ea), we need to use the Arrhenius equation:

k = A * e^(-Ea / (R * T))

Where:
- k is the rate constant
- A is the pre-exponential factor (also known as the frequency factor)
- Ea is the activation energy
- R is the gas constant (8.314 J/(mol*K))
- T is the temperature in Kelvin (K)

In this case, we are given that the reaction rate increases by a factor of 2.6 when the temperature increases from 20 to 30 degrees Celsius.

Step 1: Convert temperatures to Kelvin
20 degrees Celsius + 273.15 = 293.15 K
30 degrees Celsius + 273.15 = 303.15 K

Step 2: Express the rate constant ratio (k2/k1) in terms of the given factor of increase.
We can write k2/k1 = 2.6

Step 3: Rearrange the Arrhenius equation to solve for Ea.
Taking the natural logarithm (ln) of both sides of the equation gives:

ln(k2/k1) = (-Ea / (R * T1)) + (-Ea / (R * T2))

Since ln(a/b) = ln(a) - ln(b), we can write:

ln(k2) - ln(k1) = (-Ea / (R * T1)) + (-Ea / (R * T2))

Step 4: Substitute the values into the equation.
ln(2.6) = (-Ea / (8.314 J/(mol*K) * 293.15 K)) + (-Ea / (8.314 J/(mol*K) * 303.15 K))

Now, we can solve this equation for Ea. Use a calculator to evaluate the natural logarithm of 2.6 and solve for Ea.

Ea ā‰ˆ [(-8.314 J/(mol*K) * 293.15 K) * (-8.314 J/(mol*K) * 303.15 K)] / (8.314 J/(mol*K) * (293.15 K - 303.15 K) * ln(2.6))

By plugging in the values, you can find the approximate value of Ea.