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The t1/2 of a reaction is doubled as the initial concentration of a reactant is increased 4 times. The order of the reaction is:
- a)0.5
- b)1
- c)0
- d)3
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The t1/2 of a reaction is doubled as the initial concentration of a re...
Correct Answer :- c
Explanation : A → P . For zero order reaction, rate= k. So, unit of k is
moles/sec.
- d[A]/dt = k
Upon integration between limits, [ A ] − [ A0 ] = − kt
At t1/2, [A]=[A0]/2.
Hence, t1/2 = [A0]/2k
Increased by 4 times, so it will become double
Most Upvoted Answer
The t1/2 of a reaction is doubled as the initial concentration of a re...
**Explanation:**
The half-life (t1/2) of a reaction is the time taken for the concentration of a reactant to decrease by half. It is a measure of the rate of the reaction.
In this question, we are given that the t1/2 of a reaction is doubled when the initial concentration of a reactant is increased by 4 times.
Let's assume the initial concentration of the reactant is [A]0 and its t1/2 is t1/2.
When the initial concentration is increased by 4 times, the new initial concentration becomes 4[A]0.
According to the question, the t1/2 of the reaction is now doubled. So, the new t1/2 is 2t1/2.
We can use the integrated rate law for first-order reactions to relate the concentration of a reactant to time. The integrated rate law for a first-order reaction is:
ln([A]t/[A]0) = -kt
where [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration of the reactant, k is the rate constant, and t is time.
By rearranging the equation, we get:
t = (1/k)ln([A]0/[A]t)
From the equation, we can see that t is inversely proportional to k. If t is doubled, then k must be halved.
In this question, the t1/2 is doubled, which means the time taken for the concentration of the reactant to decrease by half is doubled. This indicates that the rate constant, k, is halved.
Since the rate constant, k, is related to the order of the reaction, we can conclude that the order of the reaction is 0 (zero).
Therefore, the correct answer is option 'C'.
The half-life (t1/2) of a reaction is the time taken for the concentration of a reactant to decrease by half. It is a measure of the rate of the reaction.
In this question, we are given that the t1/2 of a reaction is doubled when the initial concentration of a reactant is increased by 4 times.
Let's assume the initial concentration of the reactant is [A]0 and its t1/2 is t1/2.
When the initial concentration is increased by 4 times, the new initial concentration becomes 4[A]0.
According to the question, the t1/2 of the reaction is now doubled. So, the new t1/2 is 2t1/2.
We can use the integrated rate law for first-order reactions to relate the concentration of a reactant to time. The integrated rate law for a first-order reaction is:
ln([A]t/[A]0) = -kt
where [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration of the reactant, k is the rate constant, and t is time.
By rearranging the equation, we get:
t = (1/k)ln([A]0/[A]t)
From the equation, we can see that t is inversely proportional to k. If t is doubled, then k must be halved.
In this question, the t1/2 is doubled, which means the time taken for the concentration of the reactant to decrease by half is doubled. This indicates that the rate constant, k, is halved.
Since the rate constant, k, is related to the order of the reaction, we can conclude that the order of the reaction is 0 (zero).
Therefore, the correct answer is option 'C'.
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The t1/2 of a reaction is doubled as the initial concentration of a re...
Supose R=k[c]sqr then then when concentration is doubled R increases 4 times
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The t1/2 of a reaction is doubled as the initial concentration of a reactant is increased 4 times. The order of the reaction is:a)0.5b)1c)0d)3Correct answer is option 'C'. Can you explain this answer?
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