After how many second will the concentration of the reactant in a first order reaction be halved if the rate constant is $1.155 imes 10^{- 3}s^{- 1}$ ?

Question

After how many second will the concentration of the reactant in a first order reaction be halved if the rate constant is $1.155 	imes 10^{- 3}s^{- 1}$ ?

Dr. Abubacker · Accepted Answer

(a)
Rate constant $k = 1.155 	imes 10^{- 3}s^{- 1}$
$k = \frac{2.303}{t}\log\frac{a}{(a - x)}\ \ \ \because a = a,\ (a - x) = \frac{a}{2}$
$t_{1/2} = \frac{2.303}{t}\log\frac{a}{a/2}$
$= \frac{2.303}{1.155 	imes 10^{- 3}}log2$
$= \frac{2.303}{1.155 	imes 10^{- 3}} 	imes 0.3010$
$= \frac{0.693 	imes 10^{3}}{1.155} = 600\ s$
Or
$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{1.155 	imes 10^{- 3}}$
= 600 s

Q. After how many second will the concentration of the reactant in a first order reaction be halved if the rate constant is $1.155 \times 10^{- 3}s^{- 1}$ ?

Explanation

Reviewed by Dr. Abubacker

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