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Question
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is doubled?
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Solution
\[\ce{Reactant -> Product}\]
Rate of reaction (R) = k [A]2
If the concentration of the reactant is doubled, i.e., [A] = 2A, then the rate of the reaction would be,
R = k[2A]2
= k 4 A2
= 4 kA2
Change in rate of reaction = `"rate of reaction when concentration is doubled"/"rate of reaction originally"`
= `(4 kA^2)/(kA^2)`
= 4 : 1
When the concentration is doubled, the reaction rate increases by a factor of 4.
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