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A first order reaction is 50% completed in 1.26 × 1014 s. How much time would it take for 100% completion?
- a)1.26 × 1015 s
- b)2.52 × 1014 s
- c)2.52 × 1028 s
- d)infinite
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A first order reaction is 50% completed in 1.26 × 1014s. How muc...
The time taken for half the reaction to complete, i.e., the time in which the concentration of a reactant is reduced to half of its original value is called half-life period of the reaction. But it is impossible to perform 100% of the reaction. Whole of the substance never reacts because in every half-life, 50% of the substance reacts. Hence, time taken for 100% completion of a reaction is infinite.
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A first order reaction is 50% completed in 1.26 × 1014s. How muc...
To determine the half-life of a first-order reaction, we can use the equation:
t1/2 = (0.693 / k)
where t1/2 is the half-life and k is the rate constant.
In this case, we know that the reaction is 50% completed, which means that the concentration of the reactant has decreased by half.
Since the reaction is first-order, we can use the natural logarithm to solve for the rate constant:
ln(C0 / C) = kt
where C0 is the initial concentration, C is the concentration at time t, k is the rate constant, and t is the time.
Since the reaction is 50% completed, the concentration at time t is equal to half of the initial concentration (C = C0/2).
ln(C0 / (C0/2)) = k(1.26)
ln(2) = 1.26k
Now, we can solve for k:
k = ln(2) / 1.26
k ≈ 0.5513 min^-1
Finally, we can calculate the half-life:
t1/2 = (0.693 / k) = (0.693 / 0.5513)
t1/2 ≈ 1.258 min
Therefore, the half-life of the first-order reaction is approximately 1.258 minutes.
t1/2 = (0.693 / k)
where t1/2 is the half-life and k is the rate constant.
In this case, we know that the reaction is 50% completed, which means that the concentration of the reactant has decreased by half.
Since the reaction is first-order, we can use the natural logarithm to solve for the rate constant:
ln(C0 / C) = kt
where C0 is the initial concentration, C is the concentration at time t, k is the rate constant, and t is the time.
Since the reaction is 50% completed, the concentration at time t is equal to half of the initial concentration (C = C0/2).
ln(C0 / (C0/2)) = k(1.26)
ln(2) = 1.26k
Now, we can solve for k:
k = ln(2) / 1.26
k ≈ 0.5513 min^-1
Finally, we can calculate the half-life:
t1/2 = (0.693 / k) = (0.693 / 0.5513)
t1/2 ≈ 1.258 min
Therefore, the half-life of the first-order reaction is approximately 1.258 minutes.
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A first order reaction is 50% completed in 1.26 × 1014s. How muc...
A first order reaction never completes fully
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A first order reaction is 50% completed in 1.26 × 1014s. How much time would it take for 100% completion?a)1.26 × 1015sb)2.52 × 1014sc)2.52 × 1028sd)infiniteCorrect answer is option 'D'. Can you explain this answer?
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