contestada

Enter your answer in the provided box. Remember to enter your answer to the correct number of significant figures. For the reaction: A(g) + B(g) → AB(g) the rate is 0.23 mol/L·s, when [A]0 = [B]0 = 1.0 mol/L. If the reaction is first order in B and second order in A, what is the rate when [A]0 = 2.0 mol/L and [B]0 = 4.6 mol/L?

Respuesta :

Answer : The rate law is 4.232 mol/L.s

Explanation :

First we have to calculate the value of rate constant.

According to the question, the expression for rate law will be:

[tex]Rate=k[A]^2[B][/tex]

where,

k = rate constant

As we are given :

Rate law = 0.23 mol/L.s

Initial concentration of A = 1.0 mol/L

Initial concentration of B = 1.0 mol/L

Now put all the given values in the above rate law expression, we get:

[tex]0.23mol/L.s=k\times (1.0mol/L)^2\times (1.0mol/L)[/tex]

[tex]k=0.23M^{-2}s^{-1}[/tex]

Now we have to calculate the rate law when initial concentration of A and B is 2.0 mol/L and 4.6 mol//L respectively.

The expression for rate law will be:

[tex]Rate=k[A]^2[B][/tex]

where,

k = rate constant = [tex]0.23M^{-2}s^{-1}[/tex]

Rate law = ?

Initial concentration of A = 2.0 mol/L

Initial concentration of B = 4.6 mol/L

Now put all the given values in the above rate law expression, we get:

[tex]Rate=(0.23M^{-2}s^{-1})\times (2.0mol/L)^2\times (4.6mol/L)[/tex]

[tex]Rate=4.232mol/L.s[/tex]

Therefore, the rate law is 4.232 mol/L.s

The study of chemicals and bonds is called chemistry. When the amount of reactant and product are equal then is said to be an equilibrium state.

The correct answer is 4.232.

What is equilibrium constant?

  • The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium.
  • A state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change.

According to the question, the expression for rate law will be:

[tex]Rate =k[A]^2[B][/tex]

where,

  • k = rate constant

The data is given as follows:-

  • Rate law = 0.23 mol/L.s
  • Initial concentration of A = 1.0 mol/L
  • Initial concentration of B = 1.0 mol/L

Place all the values in the equation and solve it

[tex]0.23= k*1^2*1\\\\k= 0.23[/tex]

Now we have to calculate the rate law when the initial concentration of A and B is 2.0 mol/L and 4.6 mol//L respectively.

The expression for rate law will be:

[tex]Rate =k[A]^2[B][/tex]

where,

  • k = rate constant = 0.23
  • Initial concentration of A = 2.0 mol/L
  • Initial concentration of B = 4.6 mol/L

[tex]Rate = 0.23*2^2*4.6[/tex]

Hence, the rate will be 4.232.

For more information about the rate of the equation, refer to the link:-

https://brainly.com/question/16759172

Otras preguntas