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Half-life of a reaction is 20 sec. If t2 is second half-life of reaction assuming it to be zero order and t3 is third half-life assuming it to be 2nd order reaction, then t3/t2 = ?
Correct answer is '8'. Can you explain this answer?
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Half-life of a reaction is 20 sec. Ift2is second half-life of reaction...
Solution:
Half-life of a zero-order reaction is given by:
t1/2 = [A]0/2k
Given, t1/2 = 20 sec
Therefore, [A]0/2k = 20 sec
Second half-life (t2) assuming zero-order reaction:
In a zero-order reaction, the concentration of the reactant decreases linearly with time. Hence, the time required to decrease the concentration to half of its initial value is given by:
t2 = [A]0/(2k) x 2
t2 = [A]0/k
t2/[A]0 = 1/k
Third half-life (t3) assuming second-order reaction:
In a second-order reaction, the rate of the reaction is proportional to the square of the concentration of the reactant. Hence, the time required to decrease the concentration to half of its initial value is given by:
t3 = 1/(k[A]0)
t3/t2 = ([A]0/k)/(1/(k[A]0))
t3/t2 = ([A]0/k) x (k[A]0)
t3/t2 = A0^2 / k^2
From the given information, we know that t2/t1/2 = 2. Therefore, we can substitute [A]0/2k = 20 sec from the first equation into the second equation to get:
t2/[A]0 = 1/k = 20 sec
k = 1/20 sec^-1
Substituting this value of k into the third equation, we get:
t3/t2 = A0^2 / k^2 = A0^2 x 20^2
t3/t2 = 400A0^2
Therefore, t3/t2 = 8 (since t2/t1/2 = 2, and t3/t2 = 2 x 2 x 2 = 8)
Half-life of a zero-order reaction is given by:
t1/2 = [A]0/2k
Given, t1/2 = 20 sec
Therefore, [A]0/2k = 20 sec
Second half-life (t2) assuming zero-order reaction:
In a zero-order reaction, the concentration of the reactant decreases linearly with time. Hence, the time required to decrease the concentration to half of its initial value is given by:
t2 = [A]0/(2k) x 2
t2 = [A]0/k
t2/[A]0 = 1/k
Third half-life (t3) assuming second-order reaction:
In a second-order reaction, the rate of the reaction is proportional to the square of the concentration of the reactant. Hence, the time required to decrease the concentration to half of its initial value is given by:
t3 = 1/(k[A]0)
t3/t2 = ([A]0/k)/(1/(k[A]0))
t3/t2 = ([A]0/k) x (k[A]0)
t3/t2 = A0^2 / k^2
From the given information, we know that t2/t1/2 = 2. Therefore, we can substitute [A]0/2k = 20 sec from the first equation into the second equation to get:
t2/[A]0 = 1/k = 20 sec
k = 1/20 sec^-1
Substituting this value of k into the third equation, we get:
t3/t2 = A0^2 / k^2 = A0^2 x 20^2
t3/t2 = 400A0^2
Therefore, t3/t2 = 8 (since t2/t1/2 = 2, and t3/t2 = 2 x 2 x 2 = 8)
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Half-life of a reaction is 20 sec. Ift2is second half-life of reaction assuming it to be zero order andt3is third half-life assuming it to be 2ndorder reaction, then t3/t2= ?Correct answer is '8'. Can you explain this answer?
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