Answer :
Answer:
0.1066 hours
Explanation:
A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.
t1/2 = ln2/k
t1/2 = ln2/6.5 h⁻¹
t1/2 = 0.1066 h
The half-life of the pesticide is 0.1066 hours.
The half-life of the breakdown reaction is 0.1066 h
The half-life of a substance is simply defined as the time taken for half of the original substance to decay.
The half-life of a first order reaction can be obtained by the following equation:
[tex]t_{1/2} = \frac{0.693}{K}[/tex]
Where:
[tex]t_{1/2}[/tex] is the half-life
K is the decay constant
With the above formula, we can obtain the half-life of the breakdown reaction as follow:
Rate constant (K) = 6.5 h¯¹
Half-life ([tex]t_{1/2}[/tex]) =.?
[tex]t_{1/2} = \frac{0.693}{K} \\\\t_{1/2} = \frac{0.693}{6.5}\\\\t_{1/2} = 0.1066 h[/tex]
Therefore, the half-life of the breakdown reaction is 0.1066 h
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