CBSE Class 12 Chemistry 2024: question 20

Hints

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Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. For a first order reaction, $t=\frac{2.303}{k}\log \frac{[R{]}_0}{[R]}$.

   For 99% completion, $[R]=0.01[R{]}_0$:
   $t_{99\%}=\frac{2.303}{k}\log 100=\frac{2.303}{k}\times 2$

   1 mark

2. For 90% completion, $[R]=0.1[R{]}_0$:
   $t_{90\%}=\frac{2.303}{k}\log 10=\frac{2.303}{k}$

   Dividing, $\frac{t_{99\%}}{t_{90\%}}=2$, so $t_{99\%}=2t_{90\%}$.

   1 mark

Answer: $t_{99\%}=2t_{90\%}$

Revise the chapter

• Chapter: Chemical Kinetics (notes, formulas, where marks are lost and every past question)