Appeared in: 2026 (3 sets)
(a) (i) A first order reaction is $20%$ complete in $10$ minutes. Calculate the time taken when $80%$ of the reaction will be completed.
[Given : $\log 5 = 0\cdot7$, $\log 4 = 0\cdot6$]
(ii) For a chemical reaction $\text{R} \rightarrow \text{P}$, variation in $\ln[\text{R}]$ vs. time (t) graph is given below :
(I) Predict the order of the reaction.
(II) Write the slope of the curve.
OR
(b) (i) For first order reaction, calculate the ratio between the time taken to complete $\frac{3}{4}^{\text{th}}$ of the reaction and the time taken to complete half of the reaction.
(ii) (I) What is meant by a complex reaction ?
(II) Give an example of a pseudo first order reaction.
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Solution (a)
(i) Calculation for Time taken for 80% completion: [2 Marks]
- Step 1 (Formula & Given Data): [0.5 Mark]
For a first-order reaction, the rate constant $k$ is given by: $$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t}$$ Given: For 20% completion, $t_1 = 10$ min, $[R]_t = 80\% \text{ of } [R]_0 = 0.8[R]_0$. - Step 2 (Mathematical Working): [1 Mark]
Calculate $k$: $$k = \frac{2.303}{10} \log \frac{100}{80} = \frac{2.303}{10} \log 1.25$$ Now, for 80% completion, $[R]_t = 20\% \text{ of } [R]_0 = 0.2[R]_0$. $$t_2 = \frac{2.303}{k} \log \frac{100}{20} = \frac{2.303}{k} \log 5$$ Substitute $k$: $$t_2 = \frac{2.303 \times 10}{2.303 \times \log 1.25} \times \log 5 = \frac{10 \times 0.7}{\log(5/4)} = \frac{7}{0.7 - 0.6}$$ - Step 3 (Final Answer): [0.5 Mark]
$$t_2 = \frac{7}{0.1} = 70 \text{ minutes}$$
(ii) Graph Analysis: [1 Mark]
- (I) Order of reaction: [0.5 Mark]
The reaction is First Order (since the plot of $\ln[R]$ vs $t$ is a straight line). - (II) Slope of the curve: [0.5 Mark]
Slope = $-k$ (where $k$ is the rate constant).
Solution (b) (OR Choice)
(i) Ratio calculation: [2 Marks]
- Step 1 (Formula): [0.5 Mark]
For 3/4th completion: $[R]_t = [R]_0 - \frac{3}{4}[R]_0 = \frac{1}{4}[R]_0$. $$t_{3/4} = \frac{2.303}{k} \log \frac{[R]_0}{[R]_0/4} = \frac{2.303}{k} \log 4$$ - Step 2 (Working): [1 Mark]
For half-life ($t_{1/2}$): $$t_{1/2} = \frac{0.693}{k} = \frac{2.303 \log 2}{k}$$ Ratio = $\frac{t_{3/4}}{t_{1/2}} = \frac{\frac{2.303}{k} \log 4}{\frac{2.303}{k} \log 2} = \frac{2 \log 2}{\log 2} = 2$ - Step 3