Question

Time required for completion of 90% of a first order reaction is ‘t’. What is the time required for completion of 99.9%of the reaction? 

• A. t
• B. 2t
• C. 3t
• D. t/2

Answer 1

The Correct Option is C

If we want to determine the time required for completion of a certain percentage of the reaction, we can use the equation: 
ln\((\frac { [A]_₀}{ [A]})\) = kt 
Now, let's consider the given situation where the time required for completion of 90% of the reaction is 't'. This implies that at time 't', the remaining fraction of the reactant is 10% (or 0.1). 
ln \((\frac { [A]_₀}{0.1 [A]})\)) = kt 
Simplifying, 
ln \(\frac {1}{0.1}\) = kt 
ln (10) = kt 
We can rewrite this equation as:
 t = \(\frac {ln(10)}{k}\)
Now, let's find the time required for completion of 99.9% of the reaction. At this point, the remaining fraction of the reactant is 0.1% (or 0.001). 
ln \((\frac { [A]_₀}{0.001 [A]_0})\) = kt 
ln \((\frac {1}{0.001})\)= kt 
ln (1000) = kt 
Again, we can rewrite this equation as: 
t' = \(\frac {ln\ (1000)}{k}\)
Comparing t and t', we see that: 
t' = \(\frac {ln\ (1000)}{k}\)= 3 * \(\frac {ln(10)}{k}\) = 3t 
Therefore, the time required for completion of 99.9% of the reaction is 3t. 
The correct answer is (C) 3t.

Answer 2

Given:
For 90% completion: \(t = \frac{\ln(10)}{k}\)

Using the relation for a first-order reaction:
\(t = \frac{\ln\left(\frac{[A]_0}{0.1 [A]_0}\right)}{k}\)
\(t' = \frac{\ln\left(\frac{[A]_0}{0.001 [A]_0}\right)}{k}\)

Simplify using logarithmic properties:
\(\ln\left(\frac{1}{0.1}\right) = \ln(10)\)
\(\ln\left(\frac{1}{0.001}\right) = \ln(1000)\)

Calculate the time t':
\(t' = \frac{\ln(1000)}{k}\)

Compare t and t'
We have \(t' = \frac{\ln(1000)}{k}\) and we know \(t = \frac{\ln(10)}{k}\)
Now, compare t' and t:

\(t' = \frac{\ln(1000)}{k}\)
\(t = \frac{\ln(10)}{k}\)

We can rewrite \(\ln(1000)\) as \(\ln(10^3) = 3 \ln(10)\).

\(t' = \frac{3 \ln(10)}{k}\)
\(t' = 3t\)

So, the correct answer is (C): 3t