Question

per day of a certain industrial waste chemical P arrives at a treatment plant settling pond with a volume of 400.m3. P is destroyed by sunlight, and once in the pond it has a half-life of 46.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 = 0.14 g/m³

Explanation:

Complete Question

19.8 g per day of a certain industrial waste chemical P arrives at a treatment plant settling pond with a volume of 400 m³. P is destroyed by sunlight, and once in the pond it has a half-life of 46 h. Calculate the equilibrium concentration of P in the pond. Round your answer to 2 significant digits.

Let the mass of chemical P in the settling pond at any time be m

Mass flowrate of chemical P into the settling pond = 19.8 g/day = (19.8/24) g/hour = 0.825 g/hour

Mass flowrate of chemical P out of the pond = 0 g/hour

Mass flowrate out of the settling pond due to sunlight (reaction) = k m

Given, half life = 46 h

With the destruction by sunlight being a first order reaction;

Rate constant = k = (ln 2)/t,

where t = half life = 46 hours

k = (0.69315/46) = 0.01507 /hr

Mass flowrate out of the settling pond due to sunlight (reaction) = k m = (0.01507m) g/hr

(Rate of change of the mass of chemical P in the settling pond) = (Mass flowrate of chemical P into the settling pond) - (Mass flowrate of chemical P out of the settling pond) - (Mass flowrate out of the settling pond due to sunlight (reaction)

(dm/dt) = 0.825 - 0 - 0.01507m

At equilibrium, the rate of change of chemical P in the settling pond = 0; that is,(dm/dt) = 0

(dm/dt) = 0.825 - 0.01507m

0 = 0.825 - 0.01507m

0.01507m = 0.825

m = (0.825/0.01507) = 54.75 g

This is the equilibrium amount of chemical P in the settling pond.

The equilibrium concentration = (m/V)

= (54.75/400) = 0.1369 g/m³

= 0.14 g/m³ to 2 s.f

Hope this Helps!!!