The rate constant for a first order reaction is 60/s.how much time will it take to reduce the initial concentration of the reaction to its 1/4th value
Answers
Answer:
Let a M be the initial concentration.
Final concentration will be
16
a
M.
Rate constant, k=60/s
t=
k
2.303
log(
[A]
[A]
0
)
t=
60
2.303
log
⎝
⎛
16
a
a
⎠
⎞
t=4.6×10
−2
seconds
Answered By me
Answer:
It would take approximately 0.078 seconds to reduce the initial concentration of the reaction to its 1/4th value.
Explanation:
For a first-order reaction, the rate law is given by the equation:
Rate = k[A]
where "k" is the rate constant and "[A]" is the concentration of the reactant.
In this case, the rate constant (k) is given as 60/s. We want to find the time it takes to reduce the initial concentration ([A]₀) to its 1/4th value ([A]₀/4).
The integrated rate law for a first-order reaction is:
ln([A]₀/[A]) = kt
where [A]₀ is the initial concentration, [A] is the concentration at time "t," and "k" is the rate constant.
We can rearrange the equation to solve for time (t):
kt = ln([A]₀/[A])
t = (1/k) * ln([A]₀/[A])
In this case, we want to find the time it takes to reduce the initial concentration to its 1/4th value:
[A]₀/4 = [A]₀ * e^(-kt)
Dividing both sides by [A]₀ and taking the natural logarithm:
ln(1/4) = -kt
Rearranging the equation:
t = -ln(1/4) / k
Substituting the given value of k (60/s) into the equation:
t = -ln(1/4) / 60
Calculating the value:
t ≈ 0.078 s
Therefore, it would take approximately 0.078 seconds to reduce the initial concentration of the reaction to its 1/4th value.
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