For a rate law of the form, Rate = `k[A]^(m)[B]^(n)`, the exponents m and n are obtained from :

For the reaction : aA+bBrarrcC , the rate law is , r=k[A]^(m)[B]^(n) . If the initial concentration of A and B are doubled, then find the ratio of the new initial rate to the original initial rate.

The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentration of the reactants raised to some powers for the general reaction. aA+bBtocC+dD Rate law takes the form r=k[A]^(x)[B]^(y) where x and y are number that must be determined experimentely k is the rate constant and [A] and [B] are concentration of A & B respectively. Gaseous reaction AtoB+C follows first order kinetics concentration of A changes from A changes from 1M to 0.25M in 138.6 min.Find the rate of reaction when conc. of A is 0.1 M.

The rate law of a reaction is : Rate = k[A]^(2) [B] On doubling the concentration of both A and B the rate x will become :

The rare law for a reaction between th substances A and B is given by Rate =k[A]^(m)[B]^(n) On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as

The rate law for a reaction is : Rate = k[A]^(n) The rate of reaction remains same on doubling the concentration of A. The value of n is :

Rate expression for xA+yBto prodcuts is Rate =k[A]^(m)[B]^(n) . Units of K with respect to A and B respectively are s^(-1) and M^(-1).s^(-1) when concentration of A and B are increased by 4 times, then