Question

Estimate the time needed for a glycine molecule (see Table 13?4 in the textbook) to diffuse a distance of 29 ?m in water at 20 ?C if its concentration varies over that distance from 1.05 mol/m3 to 0.43 mol/m3 Compare this "speed" to its rms (thermal) speed. The molecular mass of glycine is about 75 u

Answer 1

Diffusion constant D = 95 × 10^-11 m² /s

Mass of the glycine molecule is m = 75 u

= 75 × 1.66 × 10^-27

= 124.5 × 10^-27 kg

Distance through which diffusion takes place x = 29 m = 29 × 10^-6 m

Temperature of the water is T = 20°C = 293 K

Average concentration C = ( 1.05 + 0.43 ) / 2 = 1.48 / 2

= 0.74 mol/m³

Difference in the concentration C = 1.05 - 0.43 = 0.62 mol/m³

Time taken by the glycine molecule to diffuse through a distance x is given by

t = ( C * x² ) / ( C * D )

= { 0.74 × ( 29 × 10^-6 )² } / ( 0.62 × 95 × 10^-11 )

= ( 622.34 × 10^-12 ) / ( 58.9 × 10^-11 )

= 1.056 seconds

Speed of diffusion of glycine molecule V_d = Distance / time

V_d =  x / t

= ( 29 × 10^-6 ) / ( 1.056 )

= 27.46 × 10^-6 m/s

RMS speed of the glycine molecule is V_rms = 3 k T / m

V_rms = ( 3 × 1.38 × 10^-23 × 293 ) / ( 124.5 × 10^-27 )

=  { ( 1213.02 × 10⁴ ) / 124.5 }

= 97431.32

= 312.13 m/s

To compare the two speeds , we take their ratio

v_rms / v_d = 312.13 / ( 27.46 × 10^-6 )

= 11.36 × 10⁶

= 1.136 × 10⁷

( Or ) v_d / v_rms = ( 27.46 × 10^-6 ) / 312.13

= 0.87 × 10^-7

So speed of diffusion is 10⁷ times less than rms speed of the molecule.