The entropy change in the formation of gan from its

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the entropy change in the formation of GaN from its elements under standard conditions. (c) As long as Δ f G < 0, GaN is stable with respect to its elements under standard conditions, i.e. at 1 bar N 2 . It will become unstable when Δ f G > 0. The dividing line between the two is Δ f G = 0, i.e. T = 131 530 J / mol 117 . 4 J K - 1 mol - 1 = 1120 K . (d) The reasoning here is similar to the reasoning in the previous part, except that we want Δ f G = 0. The formation reaction is Ga (l) + 1 2 N 2(g) GaN (s) . The free energy of formation of GaN therefore depends on the pressure of N 2 as follows: Δ f G = Δ f G + RT ln 1 a 1 / 2 N 2 ! = - 131 530 + 117 . 4 T + (8 . 314 472 J K - 1 mol - 1 ) T ln 1 20 1 / 2 = - 131 530 + 104 . 9 T. 2
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If we set this equation equal to zero and solve for T , we get T = 131 530 J / mol 104 . 9 J K - 1 mol - 1 = 1253 K . Looking at the formation reaction, Le Chatelier’s principle would predict that increas- ing the pressure of nitrogen should shift the equilibrium toward the products, i.e. make GaN stable over a wider temperature range or, in other words, raise the decomposition temperature, which is exactly what we observe. 3
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