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A reaction is of second order with respe...

A reaction is of second order with respect to a reactant. How is the rate of reaction affected, if the concentration of the reactant is
reduced to half?

Text Solution

Verified by Experts

When concentration of A is reduced to `(1)/(2)`,
i.e. `[A] = (1)/(2) a`
Then, rate `= k ((a)/(2)) = (1)/(4) ka^(2)`
Rate of reaction becomes `(1)/(4)` times, i.e. reduced to one-fourth.
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