At 33K. N2H4 is fifty percent dissociated. Calculate the standard free

1.	At 33K. N2H4 is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.
Answer» T = 33K N₂O₄ ⇌ 2NO₂ Initial concentration 100% Concentration dissociated 50% Concentration remaining at equilibrium 50% – 100% K_eq = 100/50 = 2 ∆G° = -2.303 RT log K_eq ∆G° = -2.303 x 8.31 x 33 x log 2 ∆G° = -190.18 J mol^-1

At 33K. N2H4 is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.

Answer»

T = 33K

N₂O₄ ⇌ 2NO₂

Initial concentration 100%

Concentration dissociated 50%

Concentration remaining at equilibrium 50% – 100%

K_eq = 100/50 = 2

∆G° = -2.303 RT log K_eq

∆G° = -2.303 x 8.31 x 33 x log 2

∆G° = -190.18 J mol^-1

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At 33K. N2H4 is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.

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