The rate equation for the reaction is established experimentally as `V=K[A]^(2)[B]^(1//2)`. What is the total order ?

The reaction A +3B to 2C obeys the rate equation Rate =k [A]^(1//2) [B]^(3//2) What is the order of the reaction ?

The rate of a reaction is given as , rate = k[X]^(3//2) [Y]^(-1//2) What is the order of the reaction ? If Y is taken large in excess , write the rate equation.

A + 2B rarr C, the rate equation for this reaction is given as Rate = k[A] [B]. If the concentration of A is kept the same but that of B is doubled what will happen to the rate itelf?

What is the order of a reaction which has a rate expression rate = K[A]^(3//2)[B]^(-1)

The rate expression for reaction A(g) +B(g) rarr C(g) is rate =k[A]^(1//2)[B]^(2) . What change in rate if initial concentration of A and B increases by factor 4 and 2 respectively ?

The rate of the reaction A+B+C to Product is given by: rate =-(d[A])/(dt)=k[A]^(1//2) [B]^(1//4) [C]^0 The order of reaction is:

(a) For a reaction , A+B rarr Product the rate law is given rate = k[A]^(1)[B]^(2) .What is the order of the reaction ? (b) write the unit of rate constant k for the first order reaction

For a reaction, A+Brarr Product, the rate law is given by r=k[A]^((1)/(2))[B]^(2) . What is the order of the reaction ?