Respuesta :
Answer: The value of the equilibrium constant is 0.024
Explanation:
Initial moles of [tex]NOBr[/tex] = 0.612 mole
Volume of container = 100 L
Initial concentration of [tex]NOBr=\frac{moles}{volume}=\frac{0.612moles}{100L}=6.12\times 10^{-3}M[/tex]
equilibrium concentration of [tex]Br_2=2.12\times 10^{-3}M[/tex]
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.
The given balanced equilibrium reaction is,
[tex]NOBr(g)\rightleftharpoons NO(g)+\frac{1}{2}Br_2(g)[/tex]
at t=0 [tex]6.12\times 10^{-3}M[/tex] 0 0
At eqm. conc. [tex](6.12\times 10^{-3}-x)M[/tex] x x/2
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[NO]^2[Br_2]}{[NOBr]}[/tex]
[tex]K_c=\frac{(2x)^2\times x/2}{(6.12\times 10^{-3}-x)}[/tex]
we are given : x/2 =[tex]2.12\times 10^{-3}[/tex]
Now put all the given values in this expression, we get :
[tex]K_c=\frac{(2.12\times 10^{-3})^{\frac{1}{2}}\times (2.12\times 10^{-3})}{(6.12\times 10^{-3})-(2.12\times 10^{-3})}[/tex]
[tex]K_c=0.024[/tex]
Thus the value of the equilibrium constant is 0.024
