What is the equilibrium constant, KC if the reaction is a gas phase reaction? (Ans.: Gas: KC = 0.328 dm3/mol)

Chemistry

1 answer:

valentina_108 [34]2 years ago

6 0

Answer:

Equilibrium constant Kc = Qc = quotient of reactant(s) and product(s)

Kc = [C]x[D]y..../[A]m[B]n..... = 0.328dm3/mol, where [C]x[D]y is the product and [A]m[B]n is the reactant(Both in gaseous states)

Explanation:

When a mixture of reactants and products of a reaction reaches equilibrium at a given temperature, its reaction quotient always has the same value. This value is called the equilibrium constant (K) of the reaction at that temperature. As for the reaction quotient, when evaluated in terms of concentrations, it is noted as Kc.

That a reaction quotient always assumes the same value at equilibrium can be expressed as:

Qc (at equilibrium) = Kc =[C]x[D]y…/[A]m[B]n…

This equation is a mathematical statement of the law of mass action: When a reaction has attained equilibrium at a given temperature, the reaction quotient for the reaction always has the same value.

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Answer:

1) ΔG°r(298 K) = - 28.619 KJ/mol

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1) ΔG°r = ∑νiΔG°f,i

⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)

from literature, T = 298 K:

∴ ΔG°CO2(g) = - 394.359 KJ/mol

∴ ΔG°CO(g) = - 137.152 KJ/mol

∴ ΔG°H2(g) = 0 KJ/mol........pure substance

∴ ΔG°H2O(g) = - 228.588 KJ/mol

⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )

⇒ ΔG°r(298 K) = - 28.619 KJ/mol

2) K = e∧(-ΔG°/RT)

∴ R = 8.314 E-3 KJ/K.mol

∴ T = 298 K

⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6

⇒ ΔG°r = - RTLnK

If T (↓) ⇒ ΔG°r (↓)

assuming T = 200 K

⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)

⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol

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