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:
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