NEET 2016 Phase 1 · ChemistryIntegrated ratePrevious Year Question

The rate of a first-order reaction is $0.04 mol L^{- 1} s^{- 1}$ at $10$ s and $0.03 mol L^{- 1} s^{- 1}$ at $20$ s after initiation of the reaction. The half-life period of the reaction is:

Answer: (A) $24.1$ s. \textbf{Answer:} (A) $24.1$ s \textbf{Solution:} For a first-order reaction, rate $r = k [R]$ , so rate is proportional to concentration: $\frac{[ R ] _{1}}{[ R ] _{2}} = \frac{r _{1}}{r _{2}} = \frac{0.04}{0.03} = \frac{4}{3}$ Apply the integrated first-order law between the two times: $k = \frac{2.303}{t _{2} - t _{1}} lo g \frac{[ R ] _{1}}{[ R ] _{2}} = \frac{2.303}{20 - 10} lo g \frac{4}{3} = \frac{2.303}{10} \times 0.1249 = 0.02877 s^{- 1}$ Then $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.02877} \approx 24.1 s$

A. $24.1$ s✓
B. $34.1$ s
C. $44.1$ s
D. $54.1$ s

Correct Answer

(A) $24.1$ s

Solution & Explanation

\textbf{Answer:} (A) $24.1$ s \textbf{Solution:} For a first-order reaction, rate $r = k [R]$ , so rate is proportional to concentration: $\frac{[ R ] _{1}}{[ R ] _{2}} = \frac{r _{1}}{r _{2}} = \frac{0.04}{0.03} = \frac{4}{3}$ Apply the integrated first-order law between the two times: $k = \frac{2.303}{t _{2} - t _{1}} lo g \frac{[ R ] _{1}}{[ R ] _{2}} = \frac{2.303}{20 - 10} lo g \frac{4}{3} = \frac{2.303}{10} \times 0.1249 = 0.02877 s^{- 1}$ Then $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.02877} \approx 24.1 s$

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The rate of a first-order reaction is 0.04 mol L−1 s−1 at 10 s and 0.03 mol L−1 s−1 at 20 s after initiation of the reaction. The half-life period of the reaction is:

Solution & Explanation

More NEET PYQ solutions

The rate of a first-order reaction is $0.04 mol L^{- 1} s^{- 1}$ at $10$ s and $0.03 mol L^{- 1} s^{- 1}$ at $20$ s after initiation of the reaction. The half-life period of the reaction is: