A reaction is second order with respect to a reaction. How is the rate of reaction affected if the (a) doubled, (b) reduced to `1//2`?
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Let the concentration of the reactant be [A] = a Rate of reaction , R = `k[A]^(2)` = `ka^(2)` (i) If the concentration of the reactant is doubled i.e., [A] = 2a , then the rate of the reaction would be `R' = k(2a)^(2)` = `4ka^(2)` = 4R Therefore , the rate of the reaction would increase by 4 times (ii) If the concentration of the reactant to half , i.e., `[A] = (1)/(2) a` then the rate of the reaction would be `R' = k ((1)/(2) a)^(2)` = `(1)/(4) ka^(2)` `= (1)/(4) R` Therefore , the rate of the reaction would be reduced to `(1^(th))/(4)` .
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