$($a$)$ A spherical particle is dissolving in excess amount of liquid. The rate of dissolution is firstorder reaction in the solvent concentration. Assume $D_i$ is the initial diameter of the sphere and $D$ is the diameter at time $t.$ If the conversion $x$ is defined as the ratio of $($the dissolved volume$)$ to $($the particle initial volume$),$ where the dissolved volume can be defined as: $V_D=\frac{(D_i^3-D^3)}{6}.$ Prove that the conversion$-$time relationship is given by:

$[1-(1-x{)}^{\frac{1}{3}}]+\frac{D_i}{2D^{**}}[1-(1-x{)}^{\frac{2}{3}}]=\frac{t}{D_i},$ where: $D^{**}=\frac{2D_e}{k_r},=\frac{2k_r{C}_A}{}$

$($b$)$ Two spherical particles of same material $(Mw=40k\frac{g}{k}$gmol $)$ having same diameter $)$ are to be completely dissolved in an acidic solution with different concentrations in a large, well$-$mixed tank. The particle density is $1800k\frac{g}{m^3}.$

i$.$ Which of the particles will be completely dissolved first?

ii$.$ In briefly, discuss how you could make the particle to be dissolved faster.

iii. Estimate the dissolution rate of the particle in the medium $1$ after $3h.$

Additional information:

Medium $1$: $D_{e1}=1\mathrm{.}42{10}^{-10}\frac{m^2}{s},k_{r1}=\frac{12}{h},C_{A1}=5\mathrm{.}5$gmo$\frac{l}{L}$

Medium $2$: $D_{e2}=3\mathrm{.}43{10}^{-10}\frac{m^2}{s},k_{r2}=\frac{15}{h},C_A=9\mathrm{.}1$gmo$\frac{l}{h}40\%$