Question

Time required for 99.9% completion of a first order reaction is _____ time the time required for completion of 90% reaction.(nearest integer).

Answer 1

Correct Answer: 3

The rate constant $K$ for a first-order reaction is given by: \[ K = \frac{1}{t} \ln\left(\frac{100}{100 - \text{completion percentage}}\right) \]
For 99.9% completion:
\[ t_{99.9\%} = \frac{\ln(1000)}{K} \]
For 90% completion:
\[ t_{90\%} = \frac{\ln(10)}{K} \]
Ratio of times:
\[ \frac{t_{99.9\%}}{t_{90\%}} = \frac{\ln(1000)}{\ln(10)} = \frac{3 \ln(10)}{\ln(10)} = 3 \]

Answer 2

This problem asks us to find the ratio of the time required for 99.9% completion of a first-order reaction to the time required for 90% completion. We need to provide the answer as the nearest integer.

Concept Used:

The integrated rate law for a first-order reaction is given by:

\[ k \cdot t = \ln \left( \frac{[A]_0}{[A]_t} \right) \]

where:

• \(k\) is the rate constant.
• \(t\) is the time.
• \([A]_0\) is the initial concentration of the reactant.
• \([A]_t\) is the concentration of the reactant at time \(t\).

The percentage completion of a reaction refers to the fraction of the initial reactant that has been consumed.

If a reaction is \(x\%\) complete, then the remaining concentration is \((100-x)\%\) of the initial concentration. So, \([A]_t = \frac{100-x}{100} [A]_0\). This means \(\frac{[A]_0}{[A]_t} = \frac{100}{100-x}\).

Step-by-Step Solution:

Step 1: Express the time required for 99.9% completion in terms of the rate constant.

For 99.9% completion, \(x = 99.9\%\). The remaining concentration is \((100 - 99.9)\% = 0.1\%\) of the initial concentration.

So, \([A]_t = \frac{0.1}{100} [A]_0 = 10^{-3} [A]_0\).

Therefore, \(\frac{[A]_0}{[A]_t} = \frac{[A]_0}{10^{-3} [A]_0} = 10^3\).

Let \(t_{99.9}\) be the time for 99.9% completion. Using the first-order integrated rate law:

\[ k \cdot t_{99.9} = \ln \left( \frac{[A]_0}{[A]_{t_{99.9}}} \right) = \ln(10^3) \]

Using the property of logarithms \(\ln(a^b) = b \ln(a)\):

\[ k \cdot t_{99.9} = 3 \ln(10) \]

Step 2: Express the time required for 90% completion in terms of the rate constant.

For 90% completion, \(x = 90\%\). The remaining concentration is \((100 - 90)\% = 10\%\) of the initial concentration.

So, \([A]_t = \frac{10}{100} [A]_0 = 0.1 [A]_0 = 10^{-1} [A]_0\).

Therefore, \(\frac{[A]_0}{[A]_t} = \frac{[A]_0}{10^{-1} [A]_0} = 10\).

Let \(t_{90}\) be the time for 90% completion. Using the first-order integrated rate law:

\[ k \cdot t_{90} = \ln \left( \frac{[A]_0}{[A]_{90}} \right) = \ln(10) \]

Step 3: Find the ratio of \(t_{99.9}\) to \(t_{90}\).

We have two equations:

\[ k \cdot t_{99.9} = 3 \ln(10) \quad \cdots (1) \] \[ k \cdot t_{90} = \ln(10) \quad \cdots (2) \]

Divide equation (1) by equation (2):

\[ \frac{k \cdot t_{99.9}}{k \cdot t_{90}} = \frac{3 \ln(10)}{\ln(10)} \]

Cancel out \(k\) and \(\ln(10)\) from both sides:

\[ \frac{t_{99.9}}{t_{90}} = 3 \]

Final Computation & Result:

The ratio of the time required for 99.9% completion to the time required for 90% completion is 3.

The question asks for this ratio as a nearest integer.

The value is exactly 3, which is an integer.

The time required for 99.9% completion of a first order reaction is 3 times the time required for completion of 90% reaction (nearest integer).