Respuesta :
Answer:
The equilibrium constant of the reaction is [tex]2.08\times 10^{-3}[/tex].
Explanation:
Formula used:
[tex]\Delta G^o=\Delta H^o-T\Delta S^o[/tex]
[tex]\Delta G^o=-RT\ln K_1[/tex]
where :
[tex]\Delta G^o[/tex] = Gibbs free energy
[tex]\Delta H^o[/tex] = Enthalpy of reaction
[tex]\Delta S^o[/tex] = Entropy of reaction
R = Gas constant = [tex]8.314J/K mol[/tex]
T = Temperature in Kelvins
[tex]K_1[/tex] = equilibrium constant at T
So we have:
[tex]\Delta G^o[/tex] = ?
[tex]\Delta H^o[/tex] = -94.9 kJ/mol = -94900 J/mol
[tex]\Delta S^o[/tex] = -224.2 J/mol K
T = 549 K
[tex]\Delta G^o=-94900 J/mol-549\times ( -224.2 J/mol K)=28,185.8 J/mol[/tex]
[tex]\Delta G^o=-RT\ln K_1[/tex]
[tex]28,185.8 J/mol=-8.314J/K mol\times 549 K\times \ln K_1[/tex]
[tex]K_1=0.002080=2.08\times 10^{-3}[/tex]
The equilibrium constant of the reaction is [tex]2.08\times 10^{-3}[/tex].
Given that ΔH° = -94.9 kJ and ΔS° = -224.2 J/K at 549 K for the given reaction, the equilibrium constant is 2.07 × 10⁻³.
What is the equilibrium constant?
The equilibrium constant (K) is a number that expresses the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.
- Step 1. Write the balanced equation.
CH₂O(g) + 2 H₂(g) → CH₄(g) + H₂O(g)
- Step 2. Calculate the standard Gibbs free energy (ΔG°).
Given that ΔH° = -94.9 kJ and ΔS° = -224.2 J/K at T = 549 K, we can calculate the ΔG° using the following expression.
ΔG° = ΔH° - T × ΔS°
ΔG° = -94.9 kJ - 549 K × (-0.2242 kJ/K) = 28.2 kJ
Step 3. Calculate the equilibrium constant for the reaction.
We will use the following expression.
ΔG° = - R × T × ln K
28.2 × 10³ J/mol = - (8.314 J/mol.K) × 549 K × ln K
K = 2.07 × 10⁻³
Given that ΔH° = -94.9 kJ and ΔS° = -224.2 J/K at 549 K for the given reaction, the equilibrium constant is 2.07 × 10⁻³.
Learn more about the equilibrium constant here: https://brainly.com/question/19340344
