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?

Hint: The rate of a reaction is directly proportional to the concentration of the reactant.Formula:Rate of reaction = Rate constant × [Reactant]^xHere, x denotes the order of reaction.Step 1: When the concentration of reactant doubled.Let,The concentration of reactant [A] = aInitial rate of reaction = RFinal rate of reaction = R_1So, for a second order reaction, x=2Rate of reaction = K × [a]^2R = Ka^2If the concentration is doubled, then[A] = 2aOn substituting this value in initial rate equation, we getRate of reaction = K × [2a]^2Rate of reaction = K × 4a^2R_1 = 4Ka^2So,Initial rate R=Ka^2 - - (i)Final rate R_1=4Ka^2 – (ii)Dividing (ii) equation by (i) we get;R_1/R = 4Ka^2/Ka^2 = 4so, R_1 = 4RHence, the rate of reaction increased 4 times.Step 2: When concentration of reactant is reduced to half.The concentration of reactant becomes [A] =[1/2]aFinal rate R_2=K((1/2)a)^2 = (1/4)Ka^2 – (iii)we know R=Ka^2 – (i)Dividing (iii) equation by (ii) we get;R_2/R = (1/4)Ka^2/Ka^2 =1/4R_2=(1/4)RHence, the rate of reaction reduce to (1/4).Final answer:(i) When the concentration of reactant doubled, the rate of reaction increased 4 times (ii) When the concentration of reactant is reduced to half, the rate of reaction reduce to (1/4)