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If ΔH = -60.0kJ and ΔS = -0.200kJ/K , the reaction is spontaneous below a certain temperature. Calculate that temperature.
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A same example .just values are changed http://answers.yahoo.com/question/index?qid=20120131155558AA5gHtk
@sarah786 is correct, use the equation for Gibbs' Free Energy\[\Delta G^0 = \Delta H^0 - T*\Delta S^0\] and set \(\Delta G\) equal to zero
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