At a certain temperature the initial concentration of no was 0.400 m and that of br2 was 0.245 m . at equilibrium the concentration of nobr was found to be 0.250 m. what is the value of kc at this temperature? what is the rate of the reaction when the initial concentration of no is 0.400 m and that of br2 is 0.245 m ?

Respuesta :

1. What is the value of kc at this temperature?

First, we must write the balanced chemical equation for the process:

2NO + Br₂ ⇌ 2NOBr

The chemical reactions that occur in a closed container can reach a state of chemical equilibrium that is characterized because the concentrations of the reactants and products remain constant over time. The equilibrium constant of a chemical reaction is the value of its reaction quotient in chemical equilibrium.

The equilibrium constant (Kc) is expressed as the ratio between the molar concentrations (mol/L) of reactants and products. Its value in a chemical reaction depends on the temperature, so it must always be specified.

So, the equilibrium constant for the above reaction is,

Kc = [tex]\frac{[NOBr]^{2} }{[Br_{2} ] [NO]^{2} }[/tex]

To facilitate the visualization of the reaction, we can write it in the following way,

2NO           +           Br₂              ⇌      2NOBr

0.400 M              0.245 M                      -                   initial

0.400 M - 2x       0.245 M - x                2x                 equilibrium

Initially the concentrations of NO and Br₂ are equal to 0.400 M and 0.245 M, respectively. As the reaction doesn't go to completion, at equilibrium an amount x dissociates of the initial concentration of Br₂ and an amount equal to 2x dissociates of the initial NO (because of the stequiometric coefficients), producing 2x of NOBr, the product of the reaction.

As the question states, at equilibrium the concentration of NOBr is equal to 0.250 M. Then,

2 x = 0.250 M → x = 0.125 M

So,

     2NO                      +           Br₂                      ⇌        2NOBr

0.400 M - 2 (0.125 M)       0.245 M - 0.125 M              0.250 M      equilibrium

  0.150 M                                0.275 M                          0.250 M

Then, the equilibrium constant will be,

Kc = [tex]\frac{[NOBr]^{2} }{[Br_{2} ] [NO]^{2} }[/tex] = [tex]\frac{(0.250 M)^{2} }{(0.275 M) (0.150)^{2} }[/tex]

Kc = 1.52 M⁻¹

So, at a certain temperature, the value of Kc is equal to 1.52 M⁻¹

2. What is the rate of the reaction when the initial concentration of NO is 0.400 m and that of Br₂ is 0.245 m?

The reaction rate is defined as the amount of substance that is transformed into a given reaction per unit of volume and time.

The rate equation or reaction law is a mathematical expression used in chemical kinetics that relates the speed of a reaction to the concentration of each reactant. The rate equation for reaction above is,

v = k [NO]²[Br₂]

where v is the speed of the reaction or the reaction rate and k is the speed constant. The value of k for this reaction is equal to k = 4444 s⁻¹ M⁻².

It is worth noting that the exponents in the concentrations of the reactants in the equation of velocity are equal to their coefficients in the equations of the equations.

If [NO] = 0.400 M and [Br₂] = 0.245 M,

v = k [NO]²[Br₂] = 4444 s⁻¹ M⁻² x (0.400)² x (0.245 M) → v = 174 M s⁻¹

So, the rate of the reaction when the initial concentration of no is 0.400 M and that of Br₂ is 0.245 M is equal to v = 174 M s⁻¹