Question

For the reaction 2A + 3B $\to$ products, 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

• A. $\frac{dx}{dt}=k\left[A{]}^2\right[B{]}^2$
• B. $\frac{dx}{dt}=k\left[A\right]\left[B\right]$
• C. $\frac{dx}{dt}=k\left[A\right]0[B{]}^2$
• D. $\frac{dx}{dt}=k[B{]}^{1/2}$

Answer 1

The correct option is D $\frac{dx}{dt}=k[B{]}^{1/2}$

A is in excess, so rate does not depend on A.