Chemical Kinetics

Chemical Kinetics

Question

A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is (i) doubled (ii) reduced to half?

Answer

Let the concentration of the reactant be [A] = a

Rate of reaction, R = k [A]2 = ka2

(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

= 4R

Therefore, the rate of the reaction would increase by 4 times.

(ii) If the concentration of the reactant is reduced to half, i.e.,left square bracket A right square bracket equals 1 half a then the rate of the reaction would be
Rate space of space reaction comma space straight R space equals space straight k open parentheses 1 half straight a squared close parentheses

equals 1 fourth ka squared

equals 1 fourth straight R

 

Therefore, the rate of the reaction would be reduced to 1 to the power of th over 4

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