The rate constant for a reaction at 305.0 K is two times the rate constant at 295.0 K. The activation energy of the reaction is ______ kJ mol
Use the Arrhenius equation.
I can use the Arrhenius equation when it is straight forward, but when it is in a question like this, I start to get lost.
The part that is confusing you PROBABLY is the part about 1 rate being twice the other one. So for the ln k2/k1 part, just use k1 and 2*k1 for the numbers. OR, just make up a number like 6 for one of them, then 12 will be the other one.
The rate constant for a reaction at 305.0 K° is two times as the rate constant at 295.0 K° ,the activation Energy of the reaction is,
To determine the activation energy of the reaction, we can use the Arrhenius equation, which relates the rate constant (k) to the temperature (T) and activation energy (Ea):
k = A * exp(-Ea / (R * T))
Where:
k = rate constant
A = pre-exponential factor
Ea = activation energy
R = gas constant (8.314 J/(mol*K))
T = temperature
Given that the rate constant at 305.0 K is two times the rate constant at 295.0 K, we can use this information to set up the following equation:
k2 = 2 * k1
Let's substitute this into the Arrhenius equation for the two temperatures:
k1 = A * exp(-Ea / (R * T1))
k2 = A * exp(-Ea / (R * T2))
Now, divide the second equation by the first equation:
(k2 / k1) = (A * exp(-Ea / (R * T2))) / (A * exp(-Ea / (R * T1)))
Since the pre-exponential factor, A, cancels out, we have:
(2) = exp(-Ea / (R * T2)) / exp(-Ea / (R * T1))
Next, take the natural logarithm (ln) of both sides of the equation:
ln(2) = ln(exp(-Ea / (R * T2)) / exp(-Ea / (R * T1)))
Using the logarithmic property ln(a/b) = ln(a) - ln(b), we can simplify the equation:
ln(2) = (-Ea / (R * T2)) - (-Ea / (R * T1))
Rearranging the equation, we get:
ln(2) = (-Ea / (R * T2)) + (Ea / (R * T1))
Now, we can factor out Ea and R:
ln(2) = Ea / R * (1 / T1 - 1 / T2)
Finally, solve for the activation energy (Ea):
Ea = R * ln(2) / (1 / T1 - 1 / T2)
Substituting the values T1 = 295.0 K and T2 = 305.0 K, and using the value for the gas constant R = 8.314 J/(mol*K), we can calculate the activation energy:
Ea = 8.314 J/(mol*K) * ln(2) / (1 / 295.0 K - 1 / 305.0 K)
Converting J to kJ:
Ea = 0.008314 kJ/(mol*K) * ln(2) / (1 / 295.0 K - 1 / 305.0 K)
Evaluating the expression, we find:
Ea ≈ 45.6 kJ/mol (rounded to 2 decimal places)
Therefore, the activation energy of the reaction is approximately 45.6 kJ/mol.