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pentagon [3]
4 years ago
6

If it takes 87.0 min for the concentration of a reactant to drop to 20.0% of its initial value in a first-order reaction, what i

s the rate constant for the reaction in the units min-1?
Chemistry
1 answer:
charle [14.2K]4 years ago
4 0

Answer:

0.0185 min⁻¹

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

Given:

20.0 % of the initial values is left which means that 0.20 of [A_0] is left. So,

\frac {[A_t]}{[A_0]} = 0.20

t = 87.0 min

\frac {[A_t]}{[A_0]}=e^{-k\times t}

0.20=e^{-k\times 87.0}

<u>k = 0.0185 min⁻¹</u>

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