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.

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.

if you mean hydrogen and oxygen then water is the product

1 is the answer to this problem