For a given reaction, the rate constant doubles when the temperature is increasd from 27C to 49C What ite activation energy? Explain.
a).347 KJ/mole
b)25.5 KJ/mole
c)31.5 KJ/mole
d)52.7 KJ/mole
To determine the activation energy for a given reaction when the rate constant doubles with a temperature increase, we can make use of the Arrhenius equation:
k = Ae^(-Ea/RT)
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
We are given that the rate constant doubles when the temperature is increased. Let's denote the rate constant at 27°C as k1 and the rate constant at 49°C as k2.
From the Arrhenius equation, we can write:
k1 = A * e^(-Ea/(R * T1))
k2 = A * e^(-Ea/(R * T2))
Since we are interested in the ratio of rate constants, we can divide the two equations:
k2/k1 = (A * e^(-Ea/(R * T2))) / (A * e^(-Ea/(R * T1)))
Now we can simplify:
k2/k1 = e^(-(Ea/R) * (1/T2 - 1/T1))
Given that the rate constant doubles when the temperature is changed from 27°C (300K) to 49°C (322K), we have:
k2/k1 = 2
Plugging in the values, we get:
2 = e^(-(Ea/R) * (1/322K - 1/300K))
To solve for Ea, let's take the natural logarithm (ln) of both sides:
ln(2) = -(Ea/R) * (1/322K - 1/300K)
Now rearrange to solve for Ea:
Ea = - R * ln(2) / (1/322K - 1/300K)
Calculating this expression, the approximate value of Ea comes out to be around 31.5 KJ/mole.
Therefore, the answer is (c) 31.5 KJ/mole.
To determine the activation energy, we can use the Arrhenius equation, which relates the rate constant (k) to the temperature (T) and the 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 (in Kelvin)
We are given that the rate constant (k) doubles when the temperature is increased from 27°C to 49°C. To simplify the calculation, let's convert these temperatures to Kelvin:
27°C + 273.15 = 300.15 K
49°C + 273.15 = 322.15 K
Now, let's consider two different temperatures in the Arrhenius equation:
k1 = A * exp(-Ea / (R * T1))
k2 = A * exp(-Ea / (R * T2))
Since the rate constant (k) doubles, we can write:
2k1 = k2
Now, substitute in the temperature values:
2 * A * exp(-Ea / (R * T1)) = A * exp(-Ea / (R * T2))
Simplifying further:
2 * exp(-Ea / (R * T1)) = exp(-Ea / (R * T2))
Now, take the natural logarithm of both sides to eliminate the exponential:
ln(2) -Ea / (R * T1) = -Ea / (R * T2)
Rearranging the equation:
-ln(2) = (Ea / (R * T1)) - (Ea / (R * T2))
Factoring out Ea and rearranging:
Ea * ((1 / (R * T1)) - (1 / (R * T2))) = -ln(2)
Now, substitute the values for R, T1, and T2:
Ea * ((1 / (8.314 J/(mol*K) * 300.15 K)) - (1 / (8.314 J/(mol*K) * 322.15 K))) = -ln(2)
Simplifying:
Ea * (0.004012 - 0.003891) = -ln(2)
Ea * 0.000121 = -ln(2)
Dividing both sides by 0.000121:
Ea = -ln(2) / 0.000121
Calculating:
Ea ≈ 25.5 KJ/mol
Therefore, the correct answer is b) 25.5 KJ/mole.