Question

Ozone decomposes to form oxygen molecules and oxygen atoms, 03(g) -› 02(g) + O(g), in the upper atmosphere. The energy of activation for this reaction is 93.1 kJ/mol. At 600 K, the rate constant for this reaction is 3.37 × 103 M-1 s1. At what temperature will the rate constant be 6 times larger, or 20.2 × 103 M-1 st?

Answer 1

At approximately 513 K, the rate constant for the reaction will be 6 times larger than at 600 K.

To solve for the temperature at which the rate constant will be 6 times larger, we can use the Arrhenius equation:

k = A * e^(-Ea/RT)

Where:

k = rate constant

A = frequency factor

Ea = activation energy

R = gas constant (8.314 J/(mol*K))

T = temperature in Kelvin

We are given:

k1 = 3.37 × 10^3 M^-1 s^-1

k2 = 20.2 × 10^3 M^-1 s^-1

Ea = 93.1 kJ/mol = 93.1 * 10^3 J/mol

We need to find T2 when k2 is 6 times larger than k1.

First, we can rewrite the Arrhenius equation to solve for T:

T = -Ea / (R * ln(k/A))

Now we can set up the equation to solve for T2:

ln(k2/A) = -Ea / (R * T2)

ln(k2/k1) = -Ea / (R * T2)

Solving for T2:

T2 = -Ea / (R * ln(k2/k1))

Plugging in the given values:

T2 = -(93.1 * 10^3 J/mol) / (8.314 J/(mol*K) * ln(20.2 × 10^3 / 3.37 × 10^3))

T2 ≈ 513 K

So, at approximately 513 K, the rate constant for the reaction will be 6 times larger than at 600 K.