The reaction SO2 (g) + NO2 (g) NO (g) + SO3 (g) has an equilibrium constant (Kc)

Question

The reaction SO2 (g) + NO2 (g) NO (g) + SO3 (g) has an equilibrium constant (Kc) of 0.690 at 450°C. If 0.100 mol each of SO2 and NO2 are placed in a 500 mL vessel at this temperature, what are the concentrations of the reactants and products when equilibrium is established for this reaction?

[SO2] & [NO2] = 0.034 M and [NO] & [SO3] = 0.166 M

[SO2] & [NO2] = 0.155 M and [NO] & [SO3] = 0.045 M

[SO2] & [NO2] = 0.131 M and [NO] & [SO3] = 0.069 M

[SO2] & [NO2] = 0.109 M and [NO] & [SO3] = 0.0908 M

Explanation / Answer

Ans- Kc = [products]^p/[reactants]^r

where p and r are the corresponding stoichiometric coefficients.hence all of the coefficients are equal to 1, so Kc for the given reaction is

Kc = ([NO]*[[SO3])/[O2]*[NO2])

The shortcut here is that we are given the numerical value for Kc (0.690), so we just need to valuate Kc for the given concentrations.

(a) [SO2] & [NO2] = 0.109 M and [NO] & [SO3] = 0.0908 M

Kc = ([NO]*[[SO3])/[O2]*[NO2]) = ((0.0908 M)*(0.0908 M))/((0.109 M)*(0.109 M))

Kc = 0.6939

(b) [SO2] & [NO2] = 0.034 M and [NO] & [SO3] = 0.166 M

Kc = ([NO]*[[SO3])/[O2]*[NO2]) = ((0.166 M)*(0.166 M ))/((0.034 M )*(0.034 M ))

Kc = 23.8373

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