Question

For a first order reaction, arrange the following in increasing time required for completion: \[ A.\ 25% \] \[ B.\ 50% \] \[ C.\ 75% \] \[ D.\ 90% \] Choose the correct answer from the options given below:
• \( \mathrm{B < A < C < D} \)
• \( \mathrm{C < B < A < D} \)
• \( \mathrm{A < B < C < D} \)
• \( \mathrm{D < C < B < A} \)

• A. \( \mathrm{B < A < C < D} \)
• B. \( \mathrm{C < B < A < D} \)
• C. \( \mathrm{A < B < C < D} \)
• D. \( \mathrm{D < C < B < A} \)

Hint:

For first order reactions: \[ t \propto \log\frac{a}{a-x} \] Higher percentage completion means smaller remaining concentration, therefore more time is required.

Answer 1

The Correct Option is C

Concept: A first order reaction is a reaction whose rate depends upon the concentration of one reactant raised to the first power. The integrated rate equation for a first order reaction is: :contentReference[oaicite:0]{index=0} where:
• \(a\) = initial concentration
• \(x\) = amount reacted
• \(a-x\) = concentration remaining after time \(t\)
• \(k\) = rate constant For a first order reaction:
• Greater completion percentage requires more time.
• Time increases logarithmically with extent of reaction.

Step 1: Understand what the percentages represent. The percentages given indicate the extent of completion of the reaction.
• \(25%\) completion means \(25%\) reactant has reacted.
• \(50%\) completion means half of reactant has reacted.
• \(75%\) completion means three-fourth reactant has reacted.
• \(90%\) completion means almost entire reactant has reacted. Clearly: \[ 25% < 50% < 75% < 90% \] As reaction progresses further toward completion, more time is needed.

Step 2: Calculate time expression for each percentage. (i) For \(25%\) completion Remaining concentration: \[ a-x = 75%\,a = 0.75a \] Thus: \[ t_{25} = \frac{2.303}{k}\log\frac{a}{0.75a} \] \[ t_{25} = \frac{2.303}{k}\log\frac{1}{0.75} \] (ii) For \(50%\) completion Remaining concentration: \[ a-x = 0.5a \] Thus: \[ t_{50} = \frac{2.303}{k}\log\frac{1}{0.5} \] \[ t_{50} = \frac{2.303}{k}\log 2 \] This is the half-life expression. (iii) For \(75%\) completion Remaining concentration: \[ a-x = 0.25a \] Thus: \[ t_{75} = \frac{2.303}{k}\log\frac{1}{0.25} \] \[ t_{75} = \frac{2.303}{k}\log 4 \] (iv) For \(90%\) completion Remaining concentration: \[ a-x = 0.1a \] Thus: \[ t_{90} = \frac{2.303}{k}\log\frac{1}{0.1} \] \[ t_{90} = \frac{2.303}{k}\log 10 \]

Step 3: Compare the logarithmic values. We know: \[ \log\frac{1}{0.75} < \log 2 < \log 4 < \log 10 \] Hence: \[ t_{25} < t_{50} < t_{75} < t_{90} \] Thus: \[ \boxed{ A < B < C < D } \]

Step 4: Match with the given options. The correct arrangement is: \[ \boxed{ A < B < C < D } \] which corresponds to: \[ \boxed{(3)} \] Additional Understanding: For first order reactions:
• Complete conversion theoretically takes infinite time.
• As reaction nears completion, reactant concentration becomes very small.
• Therefore reaction rate slows down continuously. Hence higher completion percentages require significantly larger times. Final Conclusion: Increasing order of time required is: \[ \boxed{ A < B < C < D } \] Hence, the correct answer is: \[ \boxed{(3)} \]