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10 March, 23:52

The value of K at constant temperature depends on the amounts of reactants and products that are mixed together initially. A large value of K means the equilibrium position lies far to the right. For the following reaction: CaCO3 (s) ⇌ CaO (s) + CO2 (g) the [CaCO3] appears in the denominator of the equilibrium expression. For the following reaction: H2 (g) + F2 (g) ⇌ 2HF (g) the values of K and Kp are not the same. For a reaction with K >> 1, the rate of the forward reaction is greater than the rate of the reverse reaction at equilibrium.

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  1. 11 March, 00:17
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    a) K = [ CO2 (g) ]

    ⇒ the [ CaCO3 (s) ] does not appear in the denominator of the equilibrium constant, as it is a pure solid substance.

    b) Kp = K (RT) ∧Δn

    ⇒ the values of K and Kp are not the same

    c) K >> 1, The reaction has a high yield and is said to be shifted to the right. then the rate of the forward reaction is greater than the rate of the reverse reaction at equilibrium.

    Explanation:

    a) CaCO3 (s) ↔ CaO (s) + CO2 (g)

    ⇒ K = [ CO2 (g) ]

    ∴ the [ CaCO3 (s) ] does not appear in the denominator of the equilibrium constant, as it is a pure solid substance.

    b) H2 (g) + F2 (g) ↔ 2 HF (g)

    ⇒ K = [ HF (g) ] ² / [ F2 (g) ] * [ H2 (g) ]

    ⇒ Kp = PHF² / PF2 * PH2

    for ideal gas:

    PV = RTn

    ⇒ P = n/V RT = [ ] RT

    ⇒ Kp = K (RT) ∧Δn

    ⇒ the values of K and Kp are not the same.

    c) K >> 1, The reaction has a high yield and is said to be shifted to the right. then the rate of the forward reaction is greater than the rate of the reverse reaction at equilibrium.
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