A liquid element is exposed to a solute for a single contact time t_c=4 s

A liquid element is exposed to a solute for a single contact time $t_c=4\,\mathrm{s}$. The diffusivity is $D=1\times 10^{-9}\,\mathrm{m}^2/\mathrm{s}$ and the surface–bulk concentration jump is held at $\Delta C=1\,\mathrm{mol}/\mathrm{m}^3$. Higbie’s penetration model gives the instantaneous interfacial flux \[ N(t)=\Delta C\sqrt{\frac{D}{\pi t}}\qquad(t>0). \] 1. Define the contact-averaged mass-transfer coefficient by \[ k_{L,\mathrm{avg}} :=\frac1{t_c}\int_0^{t_c}\frac{N(t)}{\Delta C}\,dt. \] Prove the exact identity \[ k_{L,\mathrm{avg}}=2\sqrt{\frac{D}{\pi t_c}} =1.784124116\times 10^{-5}\,\mathrm{m}/\mathrm{s}. \] 2. Prove that the molar uptake per unit area over the contact is \[ \int_0^{t_c}N(t)\,dt =2\Delta C\sqrt{\frac{D t_c}{\pi}} =7.136496465\times 10^{-5}\,\mathrm{mol}/\mathrm{m}^2. \] Confirm the consistency check $k_{L,\mathrm{avg}}\,\Delta C\, t_c$ equals that uptake. 3. Taking the penetration depth as the Gaussian scale $\delta=2\sqrt{D t_c}$, prove \[ \delta=1.264911064\times 10^{-4}\,\mathrm{m}. \] 4. Audit both sentences: (i) “the average coefficient is the endpoint value $k_L(t_c)=\sqrt{D/(\pi t_c)}$, so the factor $2$ is optional”; (ii) “one may average the endpoints $N(0)$ and $N(t_c)$, or replace $\delta$ by $\sqrt{D t_c}$.” Record that $N(t_c)$ is exactly half the true time-averaged flux. Do not replace Higbie averaging by a Whitman two-film resistance network, by Graetz scaling, by an exact one-dimensional similarity boundary layer of Blasius type, by a matched-diffusivity momentum/heat/mass analogy, or by a thermodiffusion slab. Do not import Fluid, Heat, Continuum, PDE, or CRE substitutions.

The antiderivative of $t^{-1/2}$ is $2 t^{1/2}$. The endpoint integrand is therefore half the mean.