Answer:
It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8
Explanation:
The equation that represents a first-order kinetics is:
Ln ([A] / [A]₀] = -kt
Where [A] is actual concentration, [A]₀ is initial concentration, K is rate constant (For the given problem, 1.57x10⁷s⁻¹ and t is time.
As you want the time when you have [A] in a factor of 8 = [A] / [A]₀ = 1/8
Replacing:
Ln ([A] / [A]₀] = -kt
Ln (1/8) = -1.57x10⁷s⁻¹*t
t = 1.32x10⁻⁷s
The time taken for the concentration of reactants to fall to 1/8 of its initial value is 1.33 × 10^-7 s.
Initial concentration (Ao) = 1.06 × 10−6 M
Concentration at time t (A) = (1.06 × 10−6 M/8) = 1.325 × 10−7 M
Rate constant (k) = 1.56 × 10^7 s −1
time taken = ?
Where
lnA = lnAo - kt
t = lnAo - lnA/-k
t = ln( 1.06 × 10−6) - ln( 1.325 × 10−7)/1.56 × 10^7
t =(-13.75 ) - (- 15.84)/1.56 × 10^7
t = 1.33 × 10^-7 s
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