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In a one component second order reaction,if the concentration of the reactant is reduced to half , the rate

Option: 1

increases two times


Option: 2

increases four times


Option: 3

decreases to one half


Option: 4

decreases to one fourth


Answers (1)

best_answer

As we have learnt,

For a second order Reaction

\mathrm{Rate = k [Reactant]^2}

So when the [Reactant] is halved, rate will become one-fourths

Hence, the correct answer is Option (4)

Posted by

Gautam harsolia

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