Question

A reaction in which A and B react to form products is first order in A and third order in B. By what factor does the reaction rate change if the concentrations of both reactants are doubled simultaneously? Enter a numerical answer only.

Answer 1

By doubling the concentrations of both reactants simultaneously, the reaction rate will change by a factor of 10.

Step-by-step explanation:

A reaction in which A and B react to form products is first order in A and third order in B. By doubling the concentration of both reactants simultaneously, the reaction rate will change by a factor of 2^1 + 2^3 = 10. This is because the rate of the reaction is proportional to the concentrations of each reactant raised to their respective orders.

Answer 2

The rate increases by factor of 16.

Step-by-step explanation:

According to the law of mass action:-

The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.

Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.

The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.

Thus,

Given that:- The order w.r.t. to A is 1 and w.r.t. to B is 3.

Thus, the rate law can be written as follows:-

`r=k[A][B]^3`

If new concentration, [A'] = 2[A] and [B'] = 2[B]

So, Rate law:-
`r'=k[A'][B']^3` =
`r'=k* 2* [A]* ({2* [B]})^3=2* 8* [A][B]^3=16r`

The rate increases by factor of 16.