Question

First order reaction concentration of the reactant is reduced to one fourth of its initial value is 50 seconds.Calculate rate constant

Answer 1

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Answer 2

Answer:  $0.028 sec^{-1}$

Explanation:

Expression for rate law for first order kinetics is given by:

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant  = ?

t = time taken = 50 seconds

a = let initial amount of the reactant  = a

a - x = amount left after decay process  = $\frac{a}{4}$

Now put all the given values in above equation, we get

$k=\frac{2.303}{50}\log\frac{a}{\frac{a}{4}}$

$k=\frac{2.303}{50}\log 4$

$k=0.028sec^{-1}$

Rate constant for the reaction is $0.028 sec^{-1}$