For the following reaction at equilibrium SO3(g) + NO(g) = SO2(g) + NO2(g)It is found that [SO2] = 0.70 M and [NO] = 1.20 M. Calculate the equilibrium constant for the readction knowing that the initial concentration were [SO3] = 2.55 M and [NO] = 1.90 M.

Answer:

[tex]K=0.14[/tex]

Explanation:

Hello!

In this case, for the undergoing chemical reaction, we can write the equilibrium expression via:

[tex]K=\frac{[SO_2][NO_2]}{[SO_3][NO]}[/tex]

Whereas the equilibrium concentration of both SO3 and NO are 2.55 M and 1.90 M respectively, it means that the extent of reaction [tex]x[/tex] is:

[tex]x=1.90M-1.20M=0.7M[/tex]

Because the equilibrium expression in terms of the reaction extent is:

[tex]K=\frac{x*x}{([SO_3]_0-x)([NO]_0-x)}[/tex]

It means that the concentration of SO3, NO, SO2 and NO2 at equilibrium are:

[tex][SO_3]=2.55M-0.70M=1.85M[/tex]

[tex][NO]=1.20M[/tex]

[tex][SO_2]=0.70M[/tex]

[tex][NO_2]=0.70M[/tex]

Thus, the equilibrium constant for such reaction is:

[tex]K=\frac{0.70*0.70}{1.85*1.90}\\\\K=0.14[/tex]

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