SATHEE: Chemical Kinetics - Result Question 44

Chemical Kinetics - Result Question 44

####44. A substance ‘A’ decomposes by a first order reaction starting initially with $[A] = 2.00 m$ and after $200 m i n$ , [A] becomes $0.15 m$ . For this reaction $t_{1 / 2}$ is

(a) $53.72 m i n$

(b) $50.49 m i n$

(d) $46.45 m i n$

[1995]

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Solution:

(a) Given initial concentration $(a) = 2.00 m$ ; Time taken $(t) = 200 m i n$ and final concentration $(a - x) = 0.15 m$ . For a first order reaction, rate constant,

$k = \frac{2.303}{t} \log \frac{a}{a - x} = \frac{2.303}{200} \log \frac{2.00}{0.15}$ $= \frac{2.303}{200} \times (0.301 + 0.824) = 1.29 \times 10^{- 2} m i n^{- 1}$ .

Further

$(t_{1 / 2}) = \frac{0.693}{k} = \frac{0.693}{1.29 \times 10^{- 2}} = 53.49 m i n$ .

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Chemical Kinetics - Result Question 44

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