For a reaction `A + 2B rarr C`, rate is given by `R = K [A] [B]^(2)`. The order of reaction is:

For the reaction A + 2B rarr C , the rate is given by R = [A] [B]^2 . Thus the order of the reaction is

For a reaction A+B rarr P , the rate is given by - Rate = [A][B]^2 How is the rate of reaction affected if the concentration of B,is doubled?

The rate of reaction A+B+C to product is given by r= K[A]^(1//3)[B]^(1//2)[C]. The order of the reaction is

For a Hypothetical elementary reaction A+B rarr P ,the rate is given is Rate=K . [A]^(2) .[B] .Then A) Rate of reaction increased by "8" times if [A] and [B] are doubled B) Order of reaction is equal to 3 C) Unit of "K" is mol^(-2)lit^(2)t^(-1) D) Order w.r.to B is zero

For a reaction: A+ B rarr P , the rate law is given as: Rate = k[A][B]^2 What is the order of reaction if A is present in large excess.

For a hypothetical reaction: A+B rarr Products, the rate law is r = k[A][B]^(0) . The order of reaction is

For a hypothetical reaction: A+B rarr Products, the rate law is r = k[A][B]^(0) . The order of reaction is

For a Hypothetical elementary reaction A+B rarr P the rate is given is Rate = K*[A]^(2)*[B] .Then (A) Rate of reaction increased by 8 times if [A] and [B] are doubled (B) Order of reaction is equal to 3 (C) Unit of "K" is mol^(-2)lit^(2)t^(-1) (D) order with respect to B is zero

For a reaction A+B rarr P , the rate is given by - Rate = [A][B]^2 What is the overall order of reaction if A is present in large excess?

For a reaction: A+ B rarr P , the rate law is given as: Rate = k[A][B]^2 How is the rate of reaction affected when the concentration of B is doubled?