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