Question:hard

For a reaction between neutral molecules \(\mathrm{X}\) and \(\mathrm{Y}\) in a solution at temperature \(T\), the measured rate of reaction is equal to the rate of diffusion. Assume \(\mathrm{X}\) is stationary and \(\mathrm{Y}\) is moving. If the diameter of the molecule \(\mathrm{X}\) is five times that of \(\mathrm{Y}\), then the rate constant for the reaction is
(\(\eta\) is viscosity of the solvent; \(k\) is the Boltzmann constant)

Show Hint

Use \(k=4\pi(D_X+D_Y)(r_X+r_Y)\) with \(D=kT/6\pi\eta r\) for both molecules, and \(r_X=5r_Y\).
Updated On: Jul 20, 2026
  • \(\dfrac{8kT}{3\eta}\)
  • \(\dfrac{4kT}{\eta}\)
  • \(\dfrac{24kT}{5\eta}\)
  • \(\dfrac{15kT}{4\eta}\)
Show Solution

The Correct Option is C

Solution and Explanation

This question uses the Smoluchowski model for a diffusion-controlled bimolecular reaction, but it is easier to solve by first writing everything as a ratio to the radius of $\mathrm{Y}$, since only the ratio of the two radii is given.

The diffusion-limited rate constant for two spherical, neutral reactants is $k = 4\pi(D_X+D_Y)(r_X+r_Y)$, and each diffusion coefficient obeys the Stokes-Einstein law $D = kT/(6\pi\eta r)$. Let $r_Y = r$, so the diameter condition "diameter of X is five times that of Y" gives $r_X = 5r$.

Write both diffusion coefficients in terms of $r$:

\[ D_X = \frac{kT}{6\pi\eta(5r)} = \frac{kT}{30\pi\eta r}, \qquad D_Y = \frac{kT}{6\pi\eta r} \]

Add them by putting both over a denominator of $30\pi\eta r$:

\[ D_X+D_Y = \frac{kT}{30\pi\eta r} + \frac{5kT}{30\pi\eta r} = \frac{6kT}{30\pi\eta r} = \frac{kT}{5\pi\eta r} \]

The sum of radii is $r_X+r_Y = 5r+r = 6r$. Multiply the two pieces into the Smoluchowski formula:

\[ k = 4\pi \times \frac{kT}{5\pi\eta r}\times 6r = \frac{4\times 6}{5}\cdot\frac{kT}{\eta} = \frac{24kT}{5\eta} \]

Notice that both the $\pi$ and the $r$ cancel exactly, which is exactly why the problem only needed to tell us the RATIO of the diameters (5:1) rather than an absolute size, a useful check that the algebra is on the right track.

So the rate constant is $k = \dfrac{24kT}{5\eta}$, which is option (C). The equal-size textbook result $8kT/3\eta$ (option A) would only apply if $r_X=r_Y$, which is not the case here, and dropping $D_X$ entirely (as in option B, $4kT/\eta$) under-counts the diffusive flux since both molecules move relative to each other.

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