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At 33K , N_(2)O_(4) is fifty percent dissociated Calculate the standard free energy change at this temperature and at one atmosphere. |
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Answer» Solution :T=33 K `N_2O_4 hArr 2NO_2` Initial concentration 100% Concentration dissociated 50% Concentration REMAINING at EQUILIBRIUM 50% - 100% `K_(eq)=100/50=2` `DeltaG^0`=-2.303 RT log `K_(eq)` `DeltaG^0`=-2.303 X 8.314 x 33 x log 2 `DeltaG^0=-190.18 "J mol"^(-1)` |
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