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For the elementary gas phase reaction H + C2H4 → C2H5, the second-order rate constant varies with tePhysical Chemistry — Kinetics Chemistry Question

Simple collision theory

For the elementary gas phase reaction H + C2H4 → C2H5, the second-order rate constant varies with temperature in the following way:

8.1.

8.1 Use the data to calculate the activation energy, Ea, and the pre-exponential factor, A, for the reaction.

Model Answer

8.1 Assuming the reaction has an Arrhenius temperature dependence, a plot of ln k vs 1/T should be linear, with slope –Ea / R and intercept ln A.
Plotting these data gives a straight line with a slope of –1042.9 K and an intercept of –23.991. We therefore have:
Ea = −(R)(slope) = −8.314 ×( −1042.9) = 8663.118 J mol-1 = 8.66 kJ mol-1.
ln A = intercept = –23.991
so A = exp( – 23.991) = 3.81×10-11 cm3 molecule-1 s-1.

8.2.

The simple collision theory of bimolecular reactions yields the following expression for the rate constant:

where µ is the reduced mass of the reactants and σ is the reaction cross section.
8.2 Interpret the role of the three factors in this expression; σ, the exponential, and the square-root term.

Model Answer

8.2 i) Explanation for the simple collision theory expression for the rate constant:

The rate of a chemical reaction must obviously be proportional to the number of collisions between the reactants. The collision rate is given by

Here, vrel = (8kT/πµ)1/2 is the mean relative velocity of the collision partners and σ is the collision cross section (the effective size of one reactant as ‘viewed’ by the other). Often, σ is set equal to π(rA + rB)2, where rA and rB are the radii of reactants A and B. nA and nB are the number densities of the two reactants.
The exponential term, exp( – Eo/RT), reflects the fact that a collision will only lead to reaction if the collision energy exceeds the activation barrier.
The overall rate is therefore

and we can identify the rate constant as

8.3.

8.3 Use the answer to part (a) to estimate σ for the reaction at 400 K.

Model Answer

8.3 From part 8.1, we have A = 3.81×10-11 cm3 molecule-1s-1. We can identify A from the simple collision theory expression as

so that

The reduced mass of H and C2H4 is

giving

8.4.

8.4 Compare the value obtained with an estimate of 4.0×10–19 m2 for the collision cross section.

Model Answer

8.4 The calculated reaction cross section is around 30 times smaller than the collision cross section. This reflects the fact that not all collisions lead to reaction. Often the collision geometry and / or internal energy states of the collision partners are important in determining whether two molecules will react when they collide.

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