Question

A reaction in which A, B, and C react to form products is zero order in A, one-half order in B, and second order in C. By what factor does the reaction rate change if [A] is doubled (and the other reactant concentrations are held constant)?

Answer 1

Answer: There will be no change in rate.

Step-by-step explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

`A+B+C\rightarrow Products`

`Rate=k[A]^x[B]^y[C]^z`

k= rate constant

x = order with respect to A = 0

y = order with respect to B =
`(1)/(2)`

z= order with respect to C = 2

Thus
`Rate=k[A]^0[B]^{(1)/(2)}[C]^2`

Given : when [A] is doubled and the other reactant concentrations are held constant.

Thus the new rate law is
`Rate'=k[2A]^0[B]^{(1)/(2)}[C]^2`

`Rate'=k[2]^0[A]^0[B]^{(1)/(2)}[C]^2`

`Rate'=k[A]^0[B]^{(1)/(2)}[C]^2`
`(2^0=1)`

`Rate'=Rate`

Thus the reaction rate would not change.