The half life of a pollutant is 0.29 per year. If the decomposition of the pollutant in water is 15°C follows a first order kinetics and a local factory spills 6.5 ×10^-3 moles of this pollutant into a lake with a volume of 2500 cm^3, what will the concentration of the pollutant ve after 24 months, assuming the lake temperature remains constant at 15°C?
First, we need to calculate the rate constant (k) using the half-life given:
t1/2 = 0.693 / k
0.29 = 0.693 / k
k = 0.693 / 0.29
k = 2.39 per year
Next, we can calculate the concentration of the pollutant after 24 months (2 years):
t = 24 months = 2 years
Using the first order kinetics equation:
ln(Ct/C0) = -kt
ln(Ct/6.5 ×10^-3) = -2.39 * 2
ln(Ct/6.5 ×10^-3) = -4.78
Ct/6.5 ×10^-3 = e^(-4.78)
Ct = 6.5 ×10^-3 * e^(-4.78)
Ct = 6.5 ×10^-3 * 0.0082
Ct = 5.33 x 10^-5 moles
Now, we need to calculate the concentration in terms of volume:
Volume of lake = 2500 cm^3 = 0.0025 m^3
Concentration (c) = moles/volume = 5.33 x 10^-5 / 0.0025
Concentration = 0.02132 moles/m^3
Therefore, the concentration of the pollutant in the lake after 24 months will be 0.02132 moles/m^3.