Question

49.9 g per day of a certain industrial waste chemical P arrives at a treatment plant settling pond with a volume of 300 m^3. P is destroyed by sunlight, and once in the pond it has a half-life of 3.4 h. Calculate the equilibrium concentration of P in the pond. Round your answer to 2 significant digits.

Answer 1

Answer:

The equilibrium concentration of P in the pond is 0.034 g/m³

Explanation:

Given that mass = 49.9 g

1 day = 24 hr

mass per hours;

Mass in = (49.9 g / day) * (1 day /24 hr )

            = 2.079 g/hr

Mass out = 0

Mass out due to sunlight = k $C_{A}$ V  

Given, half life = 3.4 h

From first order reaction; k , rate constant = ln2/t, half time

                     ln 2= 0.693, V= volume

                     k = 0.693 / t half = 0.693 / 3.4 = 0.2038 hr⁻¹

putting all values in the expression  k $C_{A}$ V  ;

Mass out due to sunlight = k $C_{A}$ V  

$C_{A}$ = Mass in hr / kV

$C_{A}$ = 2.079 g/hr / ( 0.2038 hr⁻¹ x 300 m³ )

$C_{A}$ =  0.034 g/m³

Answer 2

The equilibrium concentration of P in the pond should be 0.034 g/m³.

Calculation of the  equilibrium concentration of P:

Since

mass = 49.9 g

1 day = 24 hr

Now mass per hours is

Mass in = (49.9 g / day) * (1 day /24 hr )

= 2.079 g/hr

And,

Mass out = 0

Mass out due to sunlight = kCa V  

Also, half life = 3.4 h

Now

ln 2= 0.693, V= volume

k = 0.693 / t half = 0.693 / 3.4

= 0.2038 hr⁻¹

Now

Mass out due to sunlight = k CaV  

= Mass in hr / kCaV

= 2.079 g/hr / ( 0.2038 hr⁻¹ x 300 m³ )

=  0.034 g/m³

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