What is the rate constant of 490K if the reaction 2NOCl (g) → 2NO (g) + Cl2 (g) has
an Ea of 1.00 x 102 kJ/mol and a rate constant of 0.286 L/mol·s at 500K. What is
the rate constant at 490 K?
Use the Arrhenius equation.
ln (k2/k1) = Ea(1/T1 - 1/T2)/R
Ea = 100 J/mol
k1 = 0.286 L/mol*s
T1 = 500 K
T2 = 490 K
solve for k2.
Post your work if you get stuck.
Solution
Yes
To find the rate constant at 490 K, we can use the Arrhenius equation, which relates the rate constant (k) to the activation energy (Ea) and temperature (T). The Arrhenius equation is given by:
k = A * e^(-Ea / RT)
where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant (8.314 J/mol·K), and T is the temperature in Kelvin.
We are given the rate constant (k) at 500 K, which is 0.286 L/mol·s. Let's first convert the activation energy (Ea) from kJ/mol to J/mol by multiplying it by 1000:
Ea = 1.00 x 10^2 kJ/mol * 1000 J/1 kJ = 1.00 x 10^5 J/mol
Now we substitute the given values into the Arrhenius equation:
0.286 = A * e^(-1.00 x 10^5 / (8.314 * 500))
Now we need to solve for A. Let's rearrange the equation to isolate A:
A = 0.286 / e^(-1.00 x 10^5 / (8.314 * 500))
Using a scientific calculator, evaluate the exponential expression:
e^(-1.00 x 10^5 / (8.314 * 500)) ≈ 0.962
Now substitute this value back into the equation and solve for A:
A = 0.286 / 0.962 ≈ 0.297 L/mol·s
So, the pre-exponential factor (A) is approximately 0.297 L/mol·s.
Now we can use the pre-exponential factor (A) and the activation energy (Ea) to calculate the rate constant (k) at 490 K using the same Arrhenius equation:
k = A * e^(-Ea / RT)
T = 490 K
k = 0.297 * e^(-1.00 x 10^5 / (8.314 * 490))
Calculate the exponential expression:
e^(-1.00 x 10^5 / (8.314 * 490)) ≈ 0.974
Now substitute this value back into the equation and solve for k:
k = 0.297 * 0.974 ≈ 0.289 L/mol·s
Therefore, the rate constant at 490 K is approximately 0.289 L/mol·s.