Fireflies flash at a rate that depends on the temperature. At 29oC, the average rate is 3.3 flashes every 10 seconds, whereas at 23oC, the average falls to 2.7 flashes every 10 seconds. Calculate the "energy of activation" (in kJ/mol) for the flashing process.
rate1 = 0.33/s for T1 = 29C (302 K)
rate2 = 0.27/s for T2 = 23 (296 K)
rate1 = k1(A)
rate2 = k2(A)
Calculate k1/k1 knowing rate1 and rate2. (A) is constant.
ln(k2/k1) = Ea(1/T1-1/T2)/R
Substitute and solve for Ea.
To calculate the energy of activation for the flashing process, we can use the Arrhenius equation which describes the temperature dependence of reaction rates. The equation is given by:
k = Ae^(-Ea/RT)
Where:
k = rate constant
A = pre-exponential factor
Ea = energy of activation
R = gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
First, let's convert the given temperatures to Kelvin:
29oC + 273.15 = 302.15 K
23oC + 273.15 = 296.15 K
Next, we can use the given rate values at each temperature to determine the rate constants. Let's assign these values:
k1 = 3.3 flashes/10 seconds = 0.33 flashes/second
k2 = 2.7 flashes/10 seconds = 0.27 flashes/second
We can set up two equations using the rate constants and temperatures:
k1 = A * e^(-Ea/(R * T1))
k2 = A * e^(-Ea/(R * T2))
Dividing these equations, we get:
k1/k2 = (A * e^(-Ea/(R * T1))) / (A * e^(-Ea/(R * T2)))
k1/k2 = e^(-Ea/(R * T1) + Ea/(R * T2))
k1/k2 = e^(-Ea/R * (1/T1 - 1/T2))
k1/k2 = e^(-Ea/R * (T2 - T1)/(T1 * T2))
Now, let's substitute the values:
0.33/0.27 = e^(-Ea/(8.314 J/(mol·K)) * (296.15 K - 302.15 K)/(302.15 K * 296.15 K)
Simplifying the equation:
1.2222 = e^(-Ea/(8.314 J/(mol·K)) * (-6)/(302.15 K * 296.15 K)
1.2222 = e^(Ea/(8.314 J/(mol·K))/(302.15 K * 296.15 K)
Taking the natural logarithm of both sides:
ln(1.2222) = Ea/(8.314 J/(mol·K))/(302.15 K * 296.15 K)
Now, let's solve for Ea:
Ea = ln(1.2222) * 8.314 J/(mol·K) * (302.15 K * 296.15 K)
Converting J to kJ and simplifying:
Ea = ln(1.2222) * 8.314 kJ/(mol·K) * (302.15 K * 296.15 K)
Ea ≈ 0.216 kJ/mol
Therefore, the energy of activation for the flashing process is approximately 0.216 kJ/mol.
To calculate the energy of activation for the flashing process, we can use the Arrhenius equation, which relates the rate constant (k) to the temperature (T) and the energy of activation (Ea):
k = A * exp(-Ea / (R * T))
Where:
k is the rate constant,
A is the pre-exponential factor,
Ea is the energy of activation,
R is the ideal gas constant, and
T is the temperature in Kelvin.
First, let's convert the temperatures from Celsius to Kelvin:
29oC = 29 + 273.15 K = 302.15 K
23oC = 23 + 273.15 K = 296.15 K
Next, we can calculate the rate constants for both temperatures using the given information:
(3.3 flashes / 10 seconds) = k1
(2.7 flashes / 10 seconds) = k2
Now, we have two equations:
k1 = A * exp(-Ea / (R * 302.15 K))
k2 = A * exp(-Ea / (R * 296.15 K))
Dividing these two equations, we can cancel out the pre-exponential factor (A) and rearrange the equation to solve for the energy of activation (Ea):
k1 / k2 = exp((Ea / (R * 296.15 K)) - (Ea / (R * 302.15 K)))
Taking the natural logarithm (ln) on both sides to eliminate the exponential, we get:
ln(k1 / k2) = (Ea / (R * 296.15 K)) - (Ea / (R * 302.15 K))
Rearranging the equation, we get:
ln(k1 / k2) = Ea * ((1 / (R * 296.15 K)) - (1 / (R * 302.15 K)))
Now, we substitute the known values:
R = 8.314 J/(mol*K)
Plugging in these values, we can solve for the energy of activation (Ea).