Answer:
0.0211 M concentration of pesticide will remain after 62.0 min.
Explanation:
Initial concentration of the pesticide = [tex][A_o]=0.0314 M[/tex]
Final concentration of the pesticide after t time = [tex][A]=?[/tex]
t = 62.0 minutes
Rate constant of the reaction = k = [tex]6.40\times 10^{-3} min^{-1}[/tex]
For the first order kinetics:
[tex][A]=[A]_o\times e^{-kt}[/tex]
[tex][A]=0.0314 M\times e^{-6.40\times 10^{-3} min^{-1}\times 62.0 min} [/tex]
[tex][A]=0.0211 M[/tex]
0.0211 M concentration of pesticide will remain after 62.0 min.
A pesticide undergoes first-order decomposition with a rate constant of 6.40 × 10⁻³ min⁻¹. If the starting concentration is 0.0314 M, the remaining concentration after 62.0 min is 0.211 M.
It is a chemical reaction involving only one chemical species, in which the rate of decrease of the concentration of the reactant is directly proportional to its concentration.
A pesticide decomposes with a rate constant of 6.40 × 10⁻³ min⁻¹. If the starting concentration is 0.0314 M, we can calculate the remaining concentration after 62.0 min using the following expression.
[tex][A] = [A]_0 \times e^{-k \times t} \\[A] = 0.0314 M \times e^{-6.40 \times 10^{-3}min^{-1} \times 62.0min} = 0.211 M[/tex]
where,
A pesticide undergoes first-order decomposition with a rate constant of 6.40 × 10⁻³ min⁻¹. If the starting concentration is 0.0314 M, the remaining concentration after 62.0 min is 0.211 M.
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