Question

\(t_{99.9%}\) with respect to \(t_{90%}\) for a first order reaction is:

• A. Two
• B. One
• C. Three
• D. Four

Hint:

For first-order reactions: \[ t_{90%} : t_{99%} : t_{99.9%} = 1 : 2 : 3 \] This shortcut is extremely useful in MCQs.

Answer 1

The Correct Option is C

Concept: For a first-order reaction: \[ t = \frac{2.303}{k}\log\frac{a}{a-x} \] For percentage completion: \[ \frac{a}{a-x} = \frac{100}{\text{unreacted percentage}} \]

Step 1: Calculate \(t_{90%}\) At 90% completion: \[ 10% \text{ reactant remains} \] Thus: \[ t_{90%} = \frac{2.303}{k}\log\frac{100}{10} \] \[ = \frac{2.303}{k}\log 10 \] Since: \[ \log 10 = 1 \] \[ t_{90%} = \frac{2.303}{k} \]

Step 2: Calculate \(t_{99.9%}\) At 99.9% completion: \[ 0.1% \text{ reactant remains} \] Thus: \[ t_{99.9%} = \frac{2.303}{k}\log\frac{100}{0.1} \] \[ = \frac{2.303}{k}\log 1000 \] Since: \[ \log 1000 = 3 \] \[ t_{99.9%} = \frac{2.303}{k}\times 3 \]

Step 3: Find ratio \[ \frac{t_{99.9%}}{t_{90%}} = \frac{3\left(\frac{2.303}{k}\right)}{\left(\frac{2.303}{k}\right)} \] \[ = 3 \] Final Answer: \[ \boxed{3} \]