Question

Calculate the value of ∆g for these initial partial pressures. (the partial pressure of each gas is set a 5. 0 atm at 25°c) (use ∆g = rt ln (q/k), use the k at 25°c)

Answer 1

∆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:

brainly.com/question/13414142

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