A 100 mL reaction vessel initially contains 2.60×10^-2 moles of NO and 1.30×10^-2 moles of H2. At equilibrium the concentration of NO in the vessel is 0.161M. At equilibrium the vessel also contains N2, H2O, and H2. What is the value of the equilibrium constant for Kc for the following reaction?
2H2 (g) + 2NO(g) <—> 2H2O (g) + N2 (g)

Respuesta :

Answer:

The equilibrium constant Kc for this reaction is 19.4760

Explanation:

The volume of vessel used= [tex]100[/tex] ml

Initial moles of NO= [tex]\frac{2.60}{10^2}[/tex] moles

Initial moles of H2= [tex]\frac{1.30}{10^2}[/tex] moles

Concentration of NO at equilibrium= [tex]0.161[/tex]M

[tex]Concentration(in M)=\frac{moles}{volume(in litre)}[/tex]

Moles of NO at equilibrium= [tex]0.161(\frac{100}{1000})[/tex]

                                            =[tex]\frac{1.61}{10^2}[/tex] moles

               

                    2H2 (g)        +    2NO(g) <—>    2H2O (g) +    N2 (g)

Initial          :1.3*10^-2          2.6*10^-2                0                   0        moles

Equilibrium:1.3*10^-2 - x     2.6*10^-2-x              x                   x/2     moles

∴[tex]\frac{2.60}{10^2}-x=\frac{1.61}{10^2}[/tex]

⇒[tex]x=\frac{0.99}{10^2}[/tex]

[tex]Kc=\frac{[H2O]^2[N2]}{[H2]^2[NO]^2} (volume of vesselin litre)[/tex]

Equilibrium:0.31*10^-2      1.61*10^-2          0.99*10^-2        0.495*10^-2  moles

⇒[tex]Kc=\frac{(0.0099)^2(0.00495)}{(0.0031)^2(0.0161)^2}  (0.1)[/tex]

⇒[tex]Kc=19.4760[/tex]

The equilibrium constant has been the ratio of the equilibrium concentration of products and reactants. The equilibrium constant of the reaction has been 19.47.

What is equilibrium?

The equilibrium has been given as the state in which the rate of reaction has been constant.

From the stoichiometric law, the equilibrium concentration of the reactants and the products has been given in the ICE table attached.

The equilibrium concentration for the reaction can be given as:

[tex]\rm Kc=\dfrac{[H_2O]^2[N_2]}{[H_2]^2[NO]^2}[/tex]

Substitute the values at the equilibrium in the equation:

[tex]\rm Kc=\dfrac{(0.0099)^2(0.00495)}{(0.0031)^2(0.0161)^2}\;\times\;Volume (0.1\;L)\\ Kc=19.476[/tex]

The value of equilibrium constant for the reaction has been 19.47.

Learn more about equilibrium constant, here:

https://brainly.com/question/17960050

Ver imagen shrutiagrawal1798