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A first order reaction is 20% completed in 10 minutes. Calculate the time (in minutes) taken for the reaction to go to 80% completion: [rounded up to two decimal places]
Correct answer is between '71.80,72.40'. Can you explain this answer?
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A first order reaction is 20% completed in 10 minutes. Calculate the t...
Explanation:
Given, the reaction is a first-order reaction.
The equation for first-order reaction is,
ln ([A]t/[A]0) = -kt
where [A]t is the concentration of reactant at time t, [A]0 is the initial concentration of reactant, k is the rate constant, and t is the time.
Step 1: Finding the rate constant (k)
Let's assume the initial concentration ([A]0) to be 100. Then the concentration at 20% completion would be ([A]t) = 80.
Using the given information, we can write:
ln ([A]t/[A]0) = -kt
ln (80/100) = -k(10)
-0.223 = -10k
k = 0.0223 min^-1
Step 2: Finding the time taken for 80% completion
Now, we need to find the time taken for the reaction to go from 20% to 80% completion, which means the concentration at 80% completion would be ([A]t) = 20.
Using the same equation as before, we can write:
ln ([A]t/[A]0) = -kt
ln (20/100) = -0.0223t
-1.609 = -0.0223t
t = 71.8 min (rounded up to two decimal places)
Therefore, the time taken for the reaction to go to 80% completion is 71.8 minutes (rounded up to two decimal places).
Given, the reaction is a first-order reaction.
The equation for first-order reaction is,
ln ([A]t/[A]0) = -kt
where [A]t is the concentration of reactant at time t, [A]0 is the initial concentration of reactant, k is the rate constant, and t is the time.
Step 1: Finding the rate constant (k)
Let's assume the initial concentration ([A]0) to be 100. Then the concentration at 20% completion would be ([A]t) = 80.
Using the given information, we can write:
ln ([A]t/[A]0) = -kt
ln (80/100) = -k(10)
-0.223 = -10k
k = 0.0223 min^-1
Step 2: Finding the time taken for 80% completion
Now, we need to find the time taken for the reaction to go from 20% to 80% completion, which means the concentration at 80% completion would be ([A]t) = 20.
Using the same equation as before, we can write:
ln ([A]t/[A]0) = -kt
ln (20/100) = -0.0223t
-1.609 = -0.0223t
t = 71.8 min (rounded up to two decimal places)
Therefore, the time taken for the reaction to go to 80% completion is 71.8 minutes (rounded up to two decimal places).
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