A reaction is followed and found to have a rate constant of 3.36 × 104 M-1s-1 at 344 K and a rate constant of 7.69 M-1s-1 at 219 K. Determine the activation energy for this reaction.

Question

A reaction is followed and found to have a rate constant of 3.36 × 104 M-1s-1 at 344 K and a rate constant of 7.69 M-1s-1 at 219 K. Determine the activation energy for this reaction.

BlackTom AI · Accepted Answer

To determine the activation energy (Ea) from two rate constants at two temperatures, apply the Arrhenius relationship in its linear form: ln(k2/k1) = -Ea/R * (1/T2 - 1/T1).

First, list the values:
k1 = 3.36 × 10^4 M^-1 s^-1 at T1 = 344 K
k2 = 7.69 M^-1 s^-1 at T2 = 219 K
Note the units are consistent for a bimolecular reaction, so we can use them directly in the logarithmic ratio.

Compute the temperature terms:
1/T2 = 1/219 ≈ 0.00456621 K^-1
1/T1 = 1/344 ≈ 0.00290698 K^-1
Difference: (1/T2 - 1/T1) ≈ 0.00456621 - 0.00290698 = 0.00165923 K^-1

Compute the rate constant ratio and its natural logarithm:
k2/k1 = 7.69 / 33600 ≈ 2.2899 × 10^-4
ln(k2/k1) ≈ ln(2.2899 × 10^-4) ≈ -8.383

Solve for Ea:
Ea = -R * ln(k2/k1) / (1/T2 - 1/T1)
R = 8.314 J mol^-1 K^-1
Ea ≈ -8.314 * (-8.383) / 0.00165923 ≈ 69.66 / 0.00165923 ≈ 4.20 × 10^4 J/mol ≈ 41.9–42.0 kJ/mol

Thus, Ea is about 42.0 kJ/mol.

Now evaluate each option:
- 42.0 kJ/mol: This matches the calculated value closely; it is the correct activation energy.
- 12.5 kJ/mol: Far too small given the large change in k with temperature; would imply a much smaller temperature sensitivity than observed.
- 58.2 kJ/mol: Higher than calculated; would overestimate Ea and produce a larger magnitude of ln(k2/k1) change than observed.
- 11.5 kJ/mol: Also far too small; does not align with the observed exponential dependence of k on T.
- 23.8 kJ/mol: Still noticeably lower than the computed Ea and inconsistent with the temperature dependence in the data.

In summary, the calculation using the Arrhenius relation yields Ea ≈ 41.9–42.0 kJ/mol, which corresponds to the option 42.0 kJ/mol.

CHM-1020-LD01.2025SU PS 1-Part I (Kinetics)

A reaction is followed and found to have a rate constant of 3.36 × 104 M-1s-1 at 344 K and a rate constant of 7.69 M-1s-1 at 219 K. Determine the activation energy for this reaction.

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