Question

Water supplies are often treated with chlorine as one of the processing steps in treating wastewater. Estimate the liquid diffusion coefficient of chlorine in an infinitely dilute solution of water at 289 K using the Wilke-Chang equation

Answer 1

Answer:

⇒D_AB= 1.21×10^(-9)

Explanation:

Wike chang  equation is given as:

$D_{AB}= \frac{117.3\times10^{-18}\times\(\phi\times M_B)^{0.5}\times T}{\mu\times\nu^{0.6}}$

Where

D_AB= diffusivity of chlorine in water

Φ= 2.26 for water as solvent

ν= 0.0484 for chlorine as solute

M_B = Molecular weight of water

τ= temperature=289 K

μ= viscosity = 1.1×10^{-3}

Now putting these values in the above equation we get

$D_{AB}= \frac{117.3\times10^{-18}\times\(\2.26\times18)^{0.5}\times289}{\1.1\times10^{-3}\times\0.0484^{0.6}}$

⇒D_AB= 1.21×10^(-9)

Answer 2

Answer:

$\large \boxed{2.8\times 10^{-5}\text{ cm ^{2} /s}}$  

Explanation:

The Wilke-Chang equation for the liquid diffusion coefficient is

$D = 7.4 \times 10^{-8}\left (\dfrac{T\sqrt{xM}}{\eta V^{0.6}} \right)$

where

D = diffusion coefficient in square centimetres per second

T = kelvin temperature

x = an association parameter for the solvent

M = molar mass of solvent

η = viscosity of solvent in centipoises

V = molar volume of solvent at normal boiling point in cubic centimetres per mole

Data:

T = 289 K

x = 2.6

M = 18.02 g/mol

η = 0.890 cP

V = 18.9 cm³/mol

Calculation:

$\begin{array}{rcl}D & = & 7.4 \times 10^{-8}\left (\dfrac{T\sqrt{xM}}{\eta V^{0.6}} \right)\\\\& = & 7.4 \times 10^{-8}\left (\dfrac{289\sqrt{2.6 \times 18.02}}{0.890 (18.9)^{0.6}} \right)\\\\& = & 7.4 \times 10^{-8}\left (\dfrac{289\sqrt{46.85}}{0.890\times 5.833} \right)\\\\& = & 7.4 \times 10^{-8}\left (\dfrac{289\times 6.845}{0.890\times 5.833} \right)\\\\& = & \mathbf{2.8\times 10^{-5}}\textbf{ cm ^{2} /s}\\\end{array}$

$\text{The liquid diffusion coefficient for chlorine in water is \large \boxed{\mathbf{2.8\times 10^{-5}}\textbf{ cm ^{2} /s}} }$

The published value is 1.25 × 10⁻⁵ cm²/s.