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Chemical Kinetics

Question

If half-life of a first order reaction involving reactant A is 5 min. How long will it take [A] to react 25% of its initial concentration?

Solution

Half -life of first order reaction can be given by
t_1/2 =0.693/k

for first order reaction :
$t = \frac{2.303}{k} \log \frac{a}{a - x}$

let a=100
x= 25 (given)
time =5min

applying in above equation, we get

$5 = \frac{2.303}{k} \log \frac{100}{100 - 25} 5 = \frac{2.303}{k} \log \frac{100}{75} k = \frac{2.303}{5} \log \frac{4}{3} k = 0.4606 \log 1.33 k = 0.4606 x 0.1238 k = 0.05704$

thus for half life

t_1/2 =0.693/0.05704

Ans. 12.14 min.

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Chemical Kinetics

If half-life of a first order reaction involving reactant A is 5 min. How long will it take [A] to react 25% of its initial concentration?

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For Daily Free Study Material Join wiredfaculty

Chemical Kinetics

If half-life of a first order reaction involving reactant A is 5 min. How long will it take [A] to react 25% of its initial concentration?

Some More Questions From Chemical Kinetics Chapter

A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of the reaction.

The rate law for the decomposition of N2O5 is rate = k[N2O5]. What is the significance of k in this equation?

Define activation energy of a reaction.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

The rate law for the decomposition of N₂O₅ is rate = k[N₂O₅]. What is the significance of k in this equation?