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Zorluk: OrtaEquilibrium Constant Expression and Calculations
A 3.0 mol3.0\text{ mol} sample of SO3(g)\text{SO}_3(g) is placed in a 2.0 dm32.0\text{ dm}^3 rigid vessel and allowed to reach equilibrium at a constant temperature according to the equation:
2SO3(g)⇌2SO2(g)+O2(g)2\text{SO}_3(g) \rightleftharpoons 2\text{SO}_2(g) + \text{O}_2(g)
If 2.0 moles2.0\text{ moles} of SO2(g)\text{SO}_2(g) are present at equilibrium, what is the numerical value of the equilibrium constant, KcK_c, for this reaction?
  1. 2.0 mol dm−32.0\text{ mol dm}^{-3}Cevap
  2. B
    4.0 mol dm−34.0\text{ mol dm}^{-3}
  3. C
    1.0 mol dm−31.0\text{ mol dm}^{-3}
  4. D
    0.5 mol dm−30.5\text{ mol dm}^{-3}

Cevap

2.0 mol dm−32.0\text{ mol dm}^{-3}
The equilibrium constant KcK_c is defined as the ratio of product concentrations raised to their stoichiometric coefficients to reactant concentrations raised to their coefficients. Converting equilibrium mole quantities to molar concentrations using the 2.0 dm32.0\text{ dm}^3 volume yields [SO3]=0.5 mol dm−3[\text{SO}_3] = 0.5\text{ mol dm}^{-3}, [SO2]=1.0 mol dm−3[\text{SO}_2] = 1.0\text{ mol dm}^{-3}, and [O2]=0.5 mol dm−3[\text{O}_2] = 0.5\text{ mol dm}^{-3}. Substituting these values into Kc=[SO2]2[O2][SO3]2K_c = \frac{[\text{SO}_2]^2 [\text{O}_2]}{[\text{SO}_3]^2} gives 2.0 mol dm−32.0\text{ mol dm}^{-3}.

Adım Adım Çözüm

1
Determine equilibrium amounts of reactants and products in moles
Initial n(SO3)=3.0 moln(\text{SO}_3) = 3.0\text{ mol}. At equilibrium, n(SO2)=2.0 moln(\text{SO}_2) = 2.0\text{ mol}. From the 2:2:12:2:1 stoichiometry, 2.0 mol2.0\text{ mol} of SO3\text{SO}_3 reacted, leaving n(SO3)=3.0−2.0=1.0 moln(\text{SO}_3) = 3.0 - 2.0 = 1.0\text{ mol}. The amount of O2\text{O}_2 produced is 12×2.0=1.0 mol\frac{1}{2} \times 2.0 = 1.0\text{ mol}.
Reaction stoichiometry dictates the mole changes between reactants and products.
2
Calculate equilibrium concentrations by dividing moles by volume (V=2.0 dm3V = 2.0\text{ dm}^3)
[SO3]=1.0 mol2.0 dm3=0.5 mol dm−3[\text{SO}_3] = \frac{1.0\text{ mol}}{2.0\text{ dm}^3} = 0.5\text{ mol dm}^{-3}, [SO2]=2.0 mol2.0 dm3=1.0 mol dm−3[\text{SO}_2] = \frac{2.0\text{ mol}}{2.0\text{ dm}^3} = 1.0\text{ mol dm}^{-3}, [O2]=1.0 mol2.0 dm3=0.5 mol dm−3[\text{O}_2] = \frac{1.0\text{ mol}}{2.0\text{ dm}^3} = 0.5\text{ mol dm}^{-3}.
Equilibrium constant KcK_c requires molar concentrations in mol dm−3\text{mol dm}^{-3} rather than mole amounts.
3
Substitute concentrations into the KcK_c expression
Kc=[SO2]2[O2][SO3]2=(1.0)2×0.5(0.5)2=0.50.25=2.0 mol dm−3K_c = \frac{[\text{SO}_2]^2 [\text{O}_2]}{[\text{SO}_3]^2} = \frac{(1.0)^2 \times 0.5}{(0.5)^2} = \frac{0.5}{0.25} = 2.0\text{ mol dm}^{-3}.
The KcK_c expression places products over reactants raised to their stoichiometric powers.

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