Question

A city's water supply is contaminated with a toxin at a concentration of 0.63 mg/L. For the water to be safe for drinking, the concentration of this toxin must be below 1.5 x 10-3 mg/L. Fortunately, this toxin decomposes to a safe mixture of products by first-order kinetics with a rate constant of 0.27 day–1 . How long will it take for the water to be safe to drink?

Answer 1

Answer:

Approximately 22.37 days, will it take for the water to be safe to drink.

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,  

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

k is rate constant

Given that:- k = 0.27 (day)⁻¹

$[A_0]$ = 0.63 mg/L

$[A_t]=1.5\times 10^{-3}$ mg/L

Applying in the above equation as:-

$1.5\times 10^{-3}=0.63e^{-0.27\times t}$

$63e^{-0.27t}=150\times \:10^{-3}$

$e^{-0.27t}=\frac{1}{420}$

$t=\frac{100\ln \left(420\right)}{27}=22.37$

Approximately 22.37 days, will it take for the water to be safe to drink.

Answer 2

The time can be calculated using rate law of first order kinetics, The time taken for the water to become drinkable is 22.37 days.

The time can be calculated using rate law of first order kinetics,

$\bold {[At] = [A_0 ]e^-^k^t}$

Where,

$\bold {[At]}$ = concentration at time t = $\bold {1.5x10^-^3}$mg/L

$\bold {[A_0]}$ = initial concentration = 0.63 mg/L

k  = rate constant =   0.27 (day)⁻¹

t - time = ?

Put the values in the formula and solve it for time t,

$\bold {1.5x10^-^3} = 0.63 e ^-^0^.^2^7^\times ^t}\\\\\bold {e ^-^0^.^2^7^\times ^t = \dfrac 1{420}}\\\\\bold {t = \dfrac {100\ ln(420)}{27}}\\\\\bold {t = 22.37\ days}$

Therefore, the time taken for the water to become drinkable is 22.37 days.

To know more about rate law of first order reaction,

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