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The half life of a first order reaction is 2000 years. If the concentration after 8000 years is 0.02M, then the initial concentration was:
- a)0.04M
- b)0.16M
- c)0.32M
- d)0.08M
Correct answer is option 'C'. Can you explain this answer?
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The half life of a first order reaction is 2000 years. If the concent...
= 4, (no. of half lives)
= 0.02 × 24 = 0.32M
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The half life of a first order reaction is 2000 years. If the concent...
Given information:
- Half-life of a first-order reaction = 2000 years
- Concentration of the reaction after 8000 years = 0.02 M
To find:
- Initial concentration of the reaction
Solution:
1. Determine the rate constant (k) of the reaction using the half-life:
The half-life of a first-order reaction is given by the equation:
t1/2 = (0.693/k)
where t1/2 is the half-life and k is the rate constant.
Substituting the given half-life of 2000 years:
2000 = (0.693/k)
k = 0.693/2000
k = 0.0003465 year-1
2. Use the concentration at 8000 years and the rate constant to find the initial concentration:
The concentration of a first-order reaction at any time t is given by the equation:
[A]t = [A]0 e-kt
where [A]t is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and e is the base of the natural logarithm.
Substituting the given values:
[A]8000 = 0.02 M
k = 0.0003465 year-1
t = 8000 years
0.02 = [A]0 e-(0.0003465 x 8000)
0.02 = [A]0 e-2.772
0.02 = [A]0 x 0.0629
[A]0 = 0.02/0.0629
[A]0 = 0.317 M
Therefore, the initial concentration of the reaction was 0.32 M (option C).
Note: The answer is rounded to two significant figures to match the precision of the given concentration value.
- Half-life of a first-order reaction = 2000 years
- Concentration of the reaction after 8000 years = 0.02 M
To find:
- Initial concentration of the reaction
Solution:
1. Determine the rate constant (k) of the reaction using the half-life:
The half-life of a first-order reaction is given by the equation:
t1/2 = (0.693/k)
where t1/2 is the half-life and k is the rate constant.
Substituting the given half-life of 2000 years:
2000 = (0.693/k)
k = 0.693/2000
k = 0.0003465 year-1
2. Use the concentration at 8000 years and the rate constant to find the initial concentration:
The concentration of a first-order reaction at any time t is given by the equation:
[A]t = [A]0 e-kt
where [A]t is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and e is the base of the natural logarithm.
Substituting the given values:
[A]8000 = 0.02 M
k = 0.0003465 year-1
t = 8000 years
0.02 = [A]0 e-(0.0003465 x 8000)
0.02 = [A]0 e-2.772
0.02 = [A]0 x 0.0629
[A]0 = 0.02/0.0629
[A]0 = 0.317 M
Therefore, the initial concentration of the reaction was 0.32 M (option C).
Note: The answer is rounded to two significant figures to match the precision of the given concentration value.
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The half life of a first order reaction is 2000 years. If the concentration after 8000 years is 0.02M, then the initial concentration was:a)0.04Mb)0.16Mc)0.32Md)0.08MCorrect answer is option 'C'. Can you explain this answer?
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