Question

What is the rate law for the following mechanism in terms of the overall rate constant k? Step1:Step2:A+BB+C⇌→CD(fast)(slow) Express your answer in terms of k and the necessary concentrations (e.g., k*[A]^3*[D]).

Answer 1

Answer : The rate law for the following mechanism is $Rate=k'[A][B]^2$

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.

As we are given the mechanism for the reaction :

Step 1 : $A+B\rightleftharpoons C$    (fast)

Step 2 : $B+C\rightarrow D$    (slow)

The rate law expression for overall reaction should be in terms of B and C.

As we know that the slow step is the rate determining step. So,

The slow step reaction is,

$B+C\rightarrow D$

The expression of rate law for this reaction will be,

$Rate=k[B][C]$         ............(1)

Now applying steady state approximation for B, we get:

$K=\frac{[C]}{[A][B]}$

$[C]=K\times [A][B]$     ...........(2)

Now substituting equation 2 in 1, we get:

$Rate=k[B]\times K\times [A][B]$

$Rate=K\times k[A][B]^2$

$Rate=k'[A][B]^2$

Hence, the rate law for the following mechanism is $Rate=k'[A][B]^2$

Answer 2

The rate law is defined as the amount of reactant present in the initial concentration and used up during the course of the reaction. The rate depends on the following:-

• Amount
• Pressure
• Volume

According to the question, The expression of the rate is$Rate = k[B][C]$. B and C stand for the reactant in the reaction.

The steady-state is defined as the$K= \frac{[C]}{[A][B]}$.

After solving both the equation, the value of the rate is

$Rate = k[B] * K* [A][B]\\\\Rate= K*k[A][B]^2\\\\Rate = k'[A][B]^2$

Hence, the rate of the equation is$Rate = k'[A][B]^2$.

For more information, refer to the link:-

https://brainly.com/question/14779101