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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.

Respuesta :

Answer:

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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