Question 26 of 37intermediate🔍 AnalyzeLong Answer3 marks

For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

Correct Answer

For a first order reaction, the integrated rate law is given by:

t=2.303klog⁡[R]0[R]t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}

where [R]0[R]_0 is the initial concentration and [R][R] is the concentration at time tt.

For 99% completion, 1% of the reactant remains, so [R]=[R]0100[R] = \frac{[R]_0}{100}. Substituting this into the equation:

t99=2.303klog⁡([R]0[R]0/100)=2.303klog⁡(100)=2.303k×2t_{99} = \frac{2.303}{k} \log \left( \frac{[R]_0}{[R]_0/100} \right) = \frac{2.303}{k} \log(100) = \frac{2.303}{k} \times 2

For 90% completion, 10% of the reactant remains, so [R]=[R]010[R] = \frac{[R]_0}{10}. Substituting this into the equation:

t90=2.303klog⁡([R]0[R]0/10)=2.303klog⁡(10)=2.303k×1t_{90} = \frac{2.303}{k} \log \left( \frac{[R]_0}{[R]_0/10} \right) = \frac{2.303}{k} \log(10) = \frac{2.303}{k} \times 1

Comparing the two results, t99=2×t90t_{99} = 2 \times t_{90}. Thus, the time required for 99% completion is twice the time required for 90% completion.

Exercise: EXERCISES | Q: 3.18 | (Page 27)
For More Understanding

Explanation

The solution derives the relationship between the time taken for different completion percentages of a first-order reaction using the integrated rate law provided in the context. By substituting the remaining concentration fractions (1/100 for 99% completion and 1/10 for 90% completion) into the logarithmic term, the numerical factors of 2 and 1 are obtained respectively, establishing the required ratio.

Solution Steps

  1. Step 1: Write the integrated rate law for a first order reaction: t=2.303klog⁡[R]0[R]t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}.

  2. Step 2: Calculate time for 99% completion where [R]=[R]0/100[R] = [R]_0/100, resulting in t99=2.303k×2t_{99} = \frac{2.303}{k} \times 2.

  3. Step 3: Calculate time for 90% completion where [R]=[R]0/10[R] = [R]_0/10, resulting in t90=2.303k×1t_{90} = \frac{2.303}{k} \times 1.

  4. Step 4: Compare the results to show t99=2×t90t_{99} = 2 \times t_{90}.

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