The experimental rate law for a reaction
`2A + 2B rarr` Product, is
`V prop C_(A) C_(B)^(1//2)`. If the concentration of both A and B are doubled the rate of reaction increases by a factor of

The rate expression for the reaction A (g) + B (g) rarr C (g) is rate = kC_(A)^(2)C_(B)^(1//2) . What changes in the initial concentrations of A and B will cause the rate of reaction to increase by a factor of eight?

For the reaction A + B to products, it is found that order of A is 1 and order of B is 1/2. When concentrations of both A & B are increased four times the rate will increase by a factor

In a reaction A + B to Products (i) If the initial concentration of A is doubled and B is kept constant, the rate of the reaction is doubled (ii) If the initial concentration both A and B are doubled the rate of reaction becomes eight times The rate law of the reaction is :

In a reaction A + B to Products (i) If the initial concentration of A is doubled and B is kept constant, the rate of the reaction is doubled (ii) If the initial concentration both A and B are doubled the rate of reaction becomes eight times The rate law of the reaction is :

For the reaction A + B rarr Product, it is found that the order of A is 2 and of B is 3 in the rate expression. When concentration of both is doubled, the rate will increase by: