Final answer:
To find the time required for 80% decomposition of reactant A in a first-order reaction with a given rate constant, we can apply the first-order rate law. The time calculated using the rate constant of 0.0092 s^-1 and the natural logarithm of 0.20 is approximately 184.8 seconds.
Step-by-step explanation:
The student's question pertains to calculating the time required for 80% decomposition of reactant A in a first-order reaction with a given rate constant. The rate constant for the decomposition of reactant A is provided as 0.0092 s-1. If 80% of the reactant decomposes, this means that the remaining concentration of the reactant will be 20% of the initial concentration. Applying the integrated first-order rate law:
ln([A]/[A]0) = -kt
Where:
[A] = final concentration of A
[A]0 = initial concentration of A
k = rate constant
t = time
Since we are only interested in the fraction of reactant decomposed, we can write:
ln(0.20) = -0.0092 s-1 × t
Solving for t, we get:
t = ln(0.20) / -0.0092 s-1 ≈ 184.8 s
Therefore, to two decimal places, the time required for 80% decomposition of reactant A is 184.8 seconds, which corresponds to option a).