Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process.
If k = 2.5 x 103 s-1 at -5° Cand k = 3.5 x 104 s-1 at 25° C, what is the activation energy for the decomposition?
0.73 kJ/mol
58 kJ/mol
140 kJ/mol
580 kJ/mol
I am having massive problems with chemical kinetics so any help is good.
To determine the activation energy for the decomposition, we can use the Arrhenius equation, which relates the rate constant (k) to the temperature (T) and activation energy (Ea):
k = A * e^(-Ea/RT)
Where:
k = rate constant
A = pre-exponential factor (related to the frequency of collisions)
Ea = activation energy
R = gas constant (8.314 J/mol*K)
T = temperature in Kelvin
We are given two different temperatures (-5°C and 25°C) and their corresponding rate constants at those temperatures. We can use this information to set up two separate equations and solve for the activation energy by rearranging the equation and taking the logarithm of both sides:
ln(k1/k2) = -Ea/R * (1/T1 - 1/T2)
Where:
k1 = rate constant at temperature T1 (-5°C = 268.15 K)
k2 = rate constant at temperature T2 (25°C = 298.15 K)
Plugging in the given rate constants and temperatures, we have:
ln(2.5 x 10^3/3.5 x 10^4) = -Ea/8.314 * (1/268.15 - 1/298.15)
Now we can solve for Ea:
-2.7726 = -Ea/8.314 * (0.00373961 - 0.00335634)
Simplifying further:
-2.7726 = -Ea/8.314 * 0.00038327
Dividing both sides by 0.00038327:
-2.7726/0.00038327 = -Ea/8.314
Solving for Ea:
Ea = (-2.7726/0.00038327) * 8.314
Ea ≈ 580 kJ/mol
Therefore, the activation energy for the decomposition of dinitrogen tetraoxide is approximately 580 kJ/mol. Thus, the correct answer is 580 kJ/mol.