Question:

A certain reaction has ΔH° = -20.2 kJ and ΔS° = 36.8 J/K

by Guest65117  |  earlier

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Calculate ΔG° for the reaction at 298 K.

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  1. dG = dH -TdS

    dG = -20.2 kJ - (298K)(0.0368 kJ/K)

    dG = -20.2 kJ - 10.97kJ

    your answer: dG = -31.2kJ

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