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  • In a second-order reaction. How does the rate change when the concentration of a reactant is doubled?Option: 1 The rate remains unchanged<b

In a second-order reaction. How does the rate change when the concentration of a reactant is doubled?

Option: 1

The rate remains unchanged


Option: 2

The rate doubles


Option: 3

The rate quadruples.


Option: 4

The rate decreases by half.


Answers (1)

best_answer

In a second -order reaction. the rate is directly proportional to the square of the reactant concentration. When the concentration is doubled, the rate increase by a factor of four.

\mathrm{R=k(A)^{2}}

\mathrm{A \rightarrow 2 A}
\mathrm{R^{\prime}=k[2A]^{2}=4 k[A]^{2}=4 R}
\mathrm{R^{\prime}=4 R}

Posted by

Ritika Kankaria

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