Question

A well-mixed lake of 105 m3 is initially contaminated with chemical at a concentration of 1 mol/m3. The chemical leaves by the outflow of 0.5 m3/s and it reacts with a rate constant of 10-2 h -1. What will be the chemical concentration after 1 and 10 days and when will 90% of the chemical have left the lake? Use a time unit of seconds.

Answer 1

Concentration of chemical=1mol/m3

Volume of lake=10^5m3

So amount of chemical in the lake=1mol/m3*105m3= 10^5 mol

The amount of chemical id decresing as it is leaving with the outflow and reacting with a first order kinetics.

a) The chemical is leaving @ 0.5 m3/s=0.5m3/s

So amount of chemical outflow in 1 day=0.5m3/s* 86400 s/day=43200m3/day

Chemical leaving in 1 day=43200m3/day-1 mol/m3=43200mol/day

Chemical remaining in the lake =10^5 -43200 =56800 mol/day

b) Also, amount of chemical reaction in time=1 day as per first order kinetics

-Kt=ln [A]t/[A]o

Where [A]t= concentration of chemical after time t,

[A]o=initial chemical concentration

K=rate constant=10^-2 h-1=10^-2/3600s=2.8*10^-4*10^-2=2.8*10^-6 s-1

-Kt=ln [A]t/[A]o

-(2.8*10^-6 s-1)*86400s=ln [A]t/(56800mol)

-2419.2*10^-4= ln [A]t/(56800mol)

-0.24192= ln [A]t/(56800 mol )

0.785=[A]t/(56800 mol)

[A]t=0.785 *56800 moles =44588 mol (amount after time t=1day)

As the remaining chemical composition in the lake=44588 mol/10^5 m3=0.44588 mol/m3=0.5 mol/m3

c)Similarly, chemical remaining in the lake after 10 days=56800 mol/day *10=568000 moles

Amount reacted in 10 days,

-Kt=ln [A]t/[A]o

-(2.8*10^-6 s-1)*86400s/day *10 day =ln [A]t/(568000mol)

-2419.2*10^-3= ln [A]t/(568000mol)

-2.4192= ln [A]t/(568000 mol )

0.09=[A]t/(568000 mol)

[A]t=0.09 *568000 moles =51120 mol (amount after time t=10days)

As the remaining chemical composition in the lake=51120 mol/10^5 m3=0.51 mol/m3

d) When 90% chemical have left, remaining chemical =10% *100000=10000 moles

time=t (say)

-Kt=ln [A]t/[A]o

-(2.8*10^-6 s-1)*t =ln 10000/(100000 )

-(2.8*10^-6 s-1)*t = ln 0.1

-(2.8*10^-6 s-1)*t = -2.30

T=0.821*10^6 s=821428.6 s

T=821428.6 s/86400 s/day=9.51 days