Question

1. Assume that the rate law for a reaction is rate = k[A][B]^2

a) what is the overall order of the reaction?

b) if the concentration of both A and B are doubled, how will this affect the rate of the reaction?

c) how will doubling the concentration of A, while the concentration of B is kept constant, affect the value of k (assume that temperature does not change)? how is the rate affected?

2. It is found for the reaction 2A + B ---> C that doubling the concentration of either A or B while maintaining the other concentration constant quadruples the reaction rate. Write a rate law for the reaction.

Answer 1

Answer.(a) given rate law is r =k[A][B]²-------------(1) from the given rate law order of the reaction w.r.to A is one,and the order of the reaction w.r.to B is two.so ,the overall order of the reaction is three. order=3.       

(b) if the concentration of bothA and B are the doubled,let the rate would be r1. r1=k[2A][2B]²=8xk[A][B]²=8x r.so the rate becomes 8 times the initial rate.

(c)if the concentration of A is doubled,let rate would be r2. r2 =k[2A][B]² =2xk[A][B]²=2xr. so, the rate becomes double..

'k' value does not change because temperature is constant.

2.Answer.given reaction is 2A+B-------->C let' a' be the order of the reaction w.r.to A,ans 'b' be the order of reaction w.r.to B.then rate law is r=k₁[A]^a[B]^b----(2)

case i)if the concentration of A is doubled let the rate would be r₃,becomes quadrapled.,then    r₃=k₁[2A]^a[B]^b=2^axk₁[A]^a[B]^b=r/4

                 2^axr=r/4 ,    on solving a=-2

caseii) if the concentration of B is doubled let the rate would be r₄,becomes quadrapled.,then    r₄=k₁[A]^a[2B]^b=2^bxk₁[A]^a[B]^b=r/4

2^bxr=r/4 ,    on solving b=-2

rate law for the above reaction is r=k₁[A]^-2[B]^-2 .