Respuesta :

This is an incomplete question, here is a complete question.

Consider the following reaction:

[tex]A+B+C\rightarrow D[/tex]

The rate law for this reaction is as follows:

[tex]Rate=k\times \frac{[A][C]^2}{[B]^{1/2}}[/tex]

Suppose the rate of the reaction at certain initial concentrations of A, B, and C is 1.12 × 10⁻² M/s.

What is the rate of the reaction if the concentrations of A and C are doubled and the concentration of B is tripled?

Rate 2 = ? M/s

Answer : The rate of reaction will be, [tex]5.17\times 10^{-2}M/s[/tex]

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The given chemical reaction is,

[tex]A+B+C\rightarrow D[/tex]

The expression of rate law for this reaction will be,

[tex]\text{ Initial rate}=k\times \frac{[A][C]^2}{[B]^{1/2}}[/tex]

As the concentrations of A and C are doubled and the concentration of B is tripled then the rate of reaction will be:

[tex]Rate=k\times \frac{[2A][2C]^2}{[3B]^1/2}[/tex]

[tex]Rate=4.62k\times \frac{[A][C]^2}{[B]^{1/2}}[/tex]

[tex]Rate=4.62\times \text{ Initial rate}[/tex]

Given:

Initial rate = 1.12 × 10⁻² M/s

[tex]Rate=4.62\times 1.12\times 10^{-2}M/s[/tex]

[tex]Rate=5.17\times 10^{-2}M/s[/tex]

Thus, the rate of reaction will be, [tex]5.17\times 10^{-2}M/s[/tex]