CBSE Class 12 Chemistry 2023: question 26

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. (a) Rate of reaction $=-\frac{1}{2}\frac{d[{\mathrm{N}}_2{\mathrm{O}}_5]}{dt}$ (coefficient of ${\mathrm{N}}_2{\mathrm{O}}_5$ is 2).

   ½ mark

2. $=\frac{1}{2}\times 1.4\times 10^{-3}=0.7\times 10^{-3}=7\times 10^{-4}{\text{mol L}}^{-1}{\text{s}}^{-1}$

   ½ mark

3. (b) For a first order reaction: $t=\frac{2.303}{k}\log \frac{[R{]}_0}{[R]}$

   ½ mark

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

   ½ mark

5. For 90% completion, $[R]=\frac{[R{]}_0}{10}$: $t_{90\%}=\frac{2.303}{k}\log 10=\frac{2.303}{k}\times 1$

   ½ mark

6. $\frac{t_{99\%}}{t_{90\%}}=\frac{\log 100}{\log 10}=\frac{2}{1}$, hence $t_{99\%}=2t_{90\%}$.

   ½ mark

Answer: (a) $7\times 10^{-4}$ mol L⁻¹ s⁻¹; (b) $t_{99\%}=2t_{90\%}$ proved

Revise the chapter

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