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Sagot :
∆g for these initial partial pressures is 10,403.31 KJ.
ΔG gets increasingly positive as a product gas's partial pressure is raised. ΔG becomes more negative as the partial pressure of a reactant gas increases.
∆g = RT ln (q/k)
In this equation: R = 8.314 J mol⁻¹ K⁻¹ or 0.008314 kJ mol⁻¹ K⁻¹
K = 325
If ΔG < 0, then K > Q, and the reaction must proceed to the right to reach equilibrium.
∴∆g = RT ln (q/k)
= 8.314 × 298 ln ( 5 / 325)
= 2477.57 ln 0.015
= 2477.57 × (-4.199)
= 10,403.31 KJ
Products are preferred over reactants at equilibrium if G° 0 and both the products and reactants are in their standard states. When reactants are preferred above products in equilibrium, however, if G° > 0, K 1. At equilibrium, neither reactants nor products are preferred if G° = 0, hence K = 1.
Therefore, ∆g for these initial partial pressures is 10,403.31 KJ.
Learn more about equilibrium here:
https://brainly.com/question/13414142
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