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a) At equilibrium DGreaction = 1

b) Q < Keq when DGreaction < 0

c) When Q < Keq addtional products are formed

d) When Q = Keq the concentrations of all reactant and product species are equal to each other

e) When Q < Keq the amount of reactants will increase

2007-03-01 17:07:15 · 2 answers · asked by Samantha P 1 in Science & Mathematics Chemistry

2 answers

Lancenigo di Villorba (TV), Italy

I start recalling the "Isothermal Equation of van't Hoff", that is

DeltaG = DeltaG° + R * T * SUM[nu,i * LN(a,i)]

If I denominate SUM[nu,i * LN(a,i)] = LN(Q), I newly write

DeltaG = DeltaG° + R * T * LN(Q)

On the other hand, at the chemical equilibrium I retrieve

DeltaG = 0
LN(Q) = LN(Keq)

hence DeltaG° = DeltaG - R * T * LN(Q) = 0 - R * T * LN(Keq)
Thus, I suggest to use the following expression :

DeltaG = R * T * LN(Q / Keq)

QUESTION a) FALSE. At equilibrium's conditions, if I refer to T and P values maintained then I overcome to DeltaG = 0.

QUESTION b) TRUE. Moving to the equilibrium, if Q < Keq then DeltaG = R * T * LN(Q / Keq) < 0.

QUESTION c) TRUE. Moving to the equilibrium, if Q < Keq then reactants react furtherly toward another product's amounts.

QUESTION d) FALSE. You cannot individuate mathematical criteria leading to this assumpt.

QUESTION e) TRUE. Moving to the equilibrium, if Q < Keq then reactants react sinking themselves.

I hope this helps you.

2007-03-01 21:24:58 · answer #1 · answered by Zor Prime 7 · 0 0

is it Keg? as in a keg of beer?

2007-03-01 17:10:22 · answer #2 · answered by Pooks 6 · 0 0

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