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

`\large \boxed{2.8* 10^(-5)\text{ cm$^(2)$/s}}`

Step-by-step explanation:

The Wilke-Chang equation for the liquid diffusion coefficient is

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

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

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

Answer 2

⇒D_AB= 1.21×10^(-9)

Step-by-step explanation:

Wike chang equation is given as:

`D_(AB)= (117.3*10^(-18)*\(\phi* M_B)^(0.5)* T)/(\mu*\\u^(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)= (117.3*10^(-18)*\(\2.26*18)^(0.5)*289)/(\1.1*10^(-3)*\0.0484^(0.6))`

⇒D_AB= 1.21×10^(-9)