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The rate constant of a reaction is 0.01s-1, how much time does it take for 2.4 mol L-1 concentration of reactant reduced to 0.3 mol L-1?
  • a)
    108.3s-1
  • b)
    207.9s-1
  • c)
    248.2s-1
  • d)
    164.8s-1
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The rate constant of a reaction is 0.01s-1, how much time does it take...
Given,
K = 0.01s-1
t1/2 = 0.693/0.01
t1/2 = 69.3s
[R] = [R]0/2n
2n = [R]0/[R]
2n = 2.4/0.3
2n = 8
n = 3 (number of half-lives)
For 1 half-life t1/2 = 69.3s
For 3 half-life 3t1/2 = 3 x 69.3s = 207.9s.
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Most Upvoted Answer
The rate constant of a reaction is 0.01s-1, how much time does it take...
To determine the time it takes for the reactant concentration to decrease from 2.4 mol L-1 to 0.3 mol L-1, we can use the first-order rate equation:

ln([A]t/[A]0) = -kt

Where:
[A]t is the concentration at time t
[A]0 is the initial concentration
k is the rate constant
t is the time

We can rearrange the equation to solve for t:

t = -ln([A]t/[A]0)/k

Given:
[A]t = 0.3 mol L-1
[A]0 = 2.4 mol L-1
k = 0.01 s-1

Substituting the values into the equation:

t = -ln(0.3/2.4)/0.01

Calculating the natural logarithm:

t = -ln(0.125)/0.01

Using a calculator:

t ≈ 207.9 s

Therefore, it takes approximately 207.9 seconds for the reactant concentration to decrease from 2.4 mol L-1 to 0.3 mol L-1. The correct answer is option B.
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Community Answer
The rate constant of a reaction is 0.01s-1, how much time does it take...
Given,
K = 0.01s-1
t1/2 = 0.693/0.01
t1/2 = 69.3s
[R] = [R]0/2n
2n = [R]0/[R]
2n = 2.4/0.3
2n = 8
n = 3 (number of half-lives)
For 1 half-life t1/2 = 69.3s
For 3 half-life 3t1/2 = 3 x 69.3s = 207.9s.
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