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
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