3.30%) (open system, unsteady state, change with time) Consider a lake having th

Question

3.30%) (open system, unsteady state, change with time) Consider a lake having the following properties: Lake volume 10m3 Initial concentration of contaminant C° = 1 mol/m3 Water outflow rate 0.5 m/s Water inflow rate = 0 m3/s (there is no inflow (advection) of the contaminant into the lake) 1st order reaction rate constant k = 10-2 h' (i.e., contaminant is lost by reaction) (a) Calculate the total loss of the contaminant from the lake in mol/s (it will be a number times C because C is unknown except at time 0 or Co-1 mol/m) (b) Calculate the concentration (mol/m3) of contaminant after 1 day and after 10 days. Hint: k for the 1st order loss reaction - 0.778 x 105 s1 (c) Comment on the ratio of the concentrations after 1d and 10d. Does the ratio 10? Explain. (d) Calculate the time for 90% of the contaminant to be lost from the lake.

Explanation / Answer

Solution1.a)

we have given,

Volume of the lake=105 m3 initial concentration of contaminants=1 mol/m3 ,rate constant=10-2h-1

contamination is losing by first order reaction,

Lake----------------------------------------------->Contaminant

So rate of loss of contaminant from lake=k[lake]=KC1

=(1/100)×(1/3600)×C mol/m3-s=2.7×10-6mol/m3-s

Total loss of contaminnts=2.7×10-6×105=2.91×10-4C mol/s.

Solution1.b)

FIRST ORDER Reaction equation is

Kt = 2.303 logC0/Ct

where k=rate constatnt t=time C0=initial Concentration of contaminants Ct= Concentration after t time in lake

For 1 day time

0.778×10-5×24×60×60=2.303log1/Ct

log1/Ct=(0.778×10-5×24×60×60)/2.303=3×10-1

1/Ct=(10)0.3=1.99=Ct=1/1.99=0.5 mol/m3

For 10 day time

0.778×10-5×24×60×600=2.303log1/Ct

log1/Ct=(0.778×10-5×24×60×600)/2.303=3

1/Ct=(10)3=1000=Ct=1/1000=0.001 mol/m3

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