A first order reaction has a specific reaction rate of 10^-2 s^-1 . How much time will it take for 20 g of the reactant to reduce to 5 g ? - Chemistry MCQ NEET (UG) 2017 2026 | ExamTest.live

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A first order reaction has a specific reaction rate of 1 0^{- 2} s^{- 1} . How much time will it take for 20 g of the reactant to reduce to 5 g ?

Solution & Step-by-step Explanation

For a first-order reaction: t = \frac{2.303}{k} lo g \frac{[ A ] _{0}}{[ A ] _{t}} Given: k = 1 0^{- 2} s^{- 1}, [A]_{0} = 20 g, [A]_{t} = 5 g . t = \frac{2.303}{1 0 ^{- 2}} lo g \frac{20}{5} t = 230.3 lo g 4 t = 230.3 \times 0.6020 t = 138.6 s .Alternatively, reduction from 20 g \to 10 g \to 5 g is exactly two half-lives. t_{1/2} = \frac{0.693}{1 0 ^{- 2}} = 69.3 s .Total time = 2 \times 69.3 = 138.6 s .

Try it yourself before checking the explanation above.

A first order reaction has a specific reaction rate of 1 0^{- 2} s^{- 1} . How much time will it take for 20 g of the reactant to reduce to 5 g ?

138.6 second

346.5 second

693.0 second

238.6 second

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