Dissociation of a gas A_ 2 takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300 K. A_ 2 (~g) leftharpoons 2 ~A(~g) The standard Gibbs energy of formation of the

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For the dissociation reaction A_2(g) leftharpoons 2A(g) , first calculate the standard Gibbs free energy: Δ G°_ rxn = 2(-50.832) - (-100.00) = -1.664 kJ/mol. Using Δ G° = -RT K_p : -1664 = -8 × 300 × K_p , giving K_p = e^ 0.6933 = 2 . For dissociation with degree of dissociation α , starting with 1 mole of A_2 at 1 bar total pressure: At equilibrium, moles are (1-α) for A_2 and 2α for A , with total (1+α) moles. K_p = (4α^2)/(1-α^2) = 2 . Solving: 2(1-α^2) = 4α^2 gives α^2 = 1/3 or α = 0.577 . Since degree of dissociation is (x × 10^ -2 )^ 1/2 : (x × 10^ -2 )^ 1/2 = 0.577 gives x × 10^ -2 = 0.333 , so x = 33.3 33 .

Dissociation of a gas A_ 2 takes place according to the following...

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Substance	\Delta G_{f}^{\circ} / kJ mol^{-1}
A_2	-100.00
A	-50.832