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.

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 = $\frac{1}{2}$

z=  order with respect to C = 2

Thus $Rate=k[A]^0[B]^{\frac{1}{2}}[C]^2$

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

Thus the new rate law is $Rate'=k[2A]^0[B]^{\frac{1}{2}}[C]^2$

$Rate'=k[2]^0[A]^0[B]^{\frac{1}{2}}[C]^2$

$Rate'=k[A]^0[B]^{\frac{1}{2}}[C]^2$ $(2^0=1)$

$Rate'=Rate$

Thus the reaction rate would not change.