For the reaction 2A + 3B `rarr` product, A is in excess and on changing the concentration of B from `0.1` M to `0.4` M, rate becomes doubled. Thus, rate law is :

In the reaction A + B rarr Product , the initial rate becomes 4 times if the concentration of A is doubled (keeping B constant) and rate becomes doubled when we double the concentration of B (keeping A constant), what is the rate law and order of the reaction?

For a reaction : A+ B rarr product, it is observed that (i) on doubling the concentration of A only, the rate of reaction is also doubled. (ii) on doubling the initial concentration of both A and B, there is change by a factor of 8 in the rate of reaction. The rate of reaction is given by

For the reaction A + B rarr products, doubling the concentration of A the rate of the reaction is doubled, but on doubling the concentration of B rate reamins unaltered. The overall order of the reaction is

For a first order reaction (A) rarr products the concentration of A changes from 0.1 M to 0.025 M in 40 minutes . The rate of reaction when the concentration of A is 0.01 M is :

The reaction : X rarr Product follows first order kinetics . In 40 minutes the concentration of X changes from 0.1 to 0.025 M. Then the rate of reaction when concentration of X is 0.01 M is

For a reaction 3A rarr 2B , following graph is obtained, calculate rate of reaction when concentration of A is 0.2M .

For the reaction A+B products, it is observed that (a) On doubling the initial concentration of A only, the rate of reaction is also doubled (b) On doubling the initial concentration of both A and B, there is a change by a factor of 8 in the rate of the reaction. The rate of this reaction is given by

The rate law for the reaction A+B to Product, is given by the expression k [A] [B] . If the concentration of B is increased from 0.1 to 0.3 mol/L, keeping the value of A at 0.1 mol/L, the rate constant will be: