Question

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)

• A. \(\dfrac{8kT}{3\eta}\)
• B. \(\dfrac{4kT}{\eta}\)
• C. \(\dfrac{24kT}{5\eta}\)
• D. \(\dfrac{15kT}{4\eta}\)

Hint:

Use \(k=4\pi(D_X+D_Y)(r_X+r_Y)\) with \(D=kT/6\pi\eta r\) for both molecules, and \(r_X=5r_Y\).

Answer 1

The Correct Option is C

Step 1: Set up the diffusion-controlled rate constant.
When a reaction between two neutral species in solution is as fast as their diffusion together, the reaction is diffusion-controlled and its rate constant is given by the Smoluchowski expression
\[ k = 4\pi (D_X + D_Y)(r_X + r_Y) \]
where \(D_X, D_Y\) are the diffusion coefficients and \(r_X, r_Y\) are the radii of \(\mathrm{X}\) and \(\mathrm{Y}\).

Step 2: Write the diffusion coefficients using the Stokes-Einstein relation.
For a sphere of radius \(r\) moving through a solvent of viscosity \(\eta\), the Stokes-Einstein relation gives
\[ D = \frac{kT}{6\pi\eta r} \]
So \(D_X = \dfrac{kT}{6\pi\eta r_X}\) and \(D_Y = \dfrac{kT}{6\pi\eta r_Y}\).

Step 3: Use the given size ratio.
The diameter of \(\mathrm{X}\) is five times that of \(\mathrm{Y}\), so the radius follows the same ratio: \(r_X = 5r_Y\).
Then \(D_X = \dfrac{kT}{6\pi\eta(5r_Y)} = \dfrac{kT}{30\pi\eta r_Y}\) and \(D_Y = \dfrac{kT}{6\pi\eta r_Y}\).
\[ D_X + D_Y = \frac{kT}{30\pi\eta r_Y} + \frac{kT}{6\pi\eta r_Y} = \frac{kT}{\pi\eta r_Y}\left(\frac{1}{30}+\frac{5}{30}\right) = \frac{kT}{\pi\eta r_Y}\cdot\frac{6}{30} = \frac{kT}{5\pi\eta r_Y} \]
Also \(r_X + r_Y = 5r_Y + r_Y = 6r_Y\).

Step 4: Substitute back into the rate constant expression.
\[ k = 4\pi (D_X+D_Y)(r_X+r_Y) = 4\pi \cdot \frac{kT}{5\pi\eta r_Y}\cdot 6r_Y \]
The factor \(\pi\) cancels with the \(\pi\) in the denominator and \(r_Y\) cancels with \(r_Y\):
\[ k = \frac{4\times 6\, kT}{5\eta} = \frac{24kT}{5\eta} \]

Step 5: Check why the other options are wrong.
Option (A), \(\dfrac{8kT}{3\eta}\), is the value you get only when \(\mathrm{X}\) and \(\mathrm{Y}\) have EQUAL radii \((r_X=r_Y=r)\); it ignores the given 5:1 size ratio.
Option (B), \(\dfrac{4kT}{\eta}\), comes from wrongly setting \(D_X=0\) (treating "\(\mathrm{X}\) stationary" as meaning it does not diffuse at all) while still using \(r_X+r_Y\); the diffusion coefficient of the pair is always the SUM \(D_X+D_Y\), so this drops a real contribution.
Option (D), \(\dfrac{15kT}{4\eta}\), does not follow from the correct algebra of Steps 3-4 and is only a plausible-looking distractor value.

Final Answer:
The diffusion-controlled rate constant works out to \[ \boxed{k=\dfrac{24kT}{5\eta}} \], option (C).