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2poclâ‚(g) ⇜ 2 pclâ‚(g) + oâ‚‚(g) â†gâ° = (2 mol)î”gfâ°(pclâ‚) – (2 mol)î”gfâ°(poclâ‚) = (2 mol)(–270 –(–502) kj/mol) â†gâ° = + 464 kj = â†hⰠ– tî”sâ° î”sâ° = 179 j/k ⇒ â†hâ° = â†gâ° + (298 k)(179 j/k) = –411 kj assuming that î”hâ° and î”sâ° do not depend on temperature, the temperature at which â†gâ° = 0 is where t = â†hâ°/î”sâ° = (–411 kj)/(0.179 j/k) = 2294 k. at that temp. keq = 1 for the reaction.