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Calculate the standard free energy change for the reaction below. P4($+ 6C12(g) → 4PC13(g) kJ AG 298.Pals) = 24.44 mol AG298.
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Gºl P4 ) = 24.44 kJ/mol Gº ( C12 )= 0.00 kJ/mol Gº ( PC13 ) = -267.8 kJ/mol 1 P4 (s) + 6 C12 (g) H 4 PC13 (g) AGºrxn = AGf pr

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