Distinguish between ΔG0', ΔG0, ΔG, and Gf0.
Gf0 => Standard Gibbs free energy for formation.
ΔG0 => Change in standard Gibbs free energy. This is the free energy at 298 K.
ΔG => ΔG = ΔG0 + RT ln K
Where ΔG is the change in free energy
R = gas constant
T = temperature of the reaction
K = equilibrium constant for the reaction.
At equilibrium, ΔG = 0
So, ΔG0 = - RT ln K
ΔG0' = Change in standard Gibbs free energy corresponding to an apparent equilibrium constant.
Suppose one of the concentration of the reactant is kept constant, and other are changed. So, equilibrium constant can be found apparent and is K'.
ΔG0' = - RT ln K'
What is the difference between ΔG and ΔG°? When does ΔG = ΔG°?
Calculate ΔG0 for the following reactions at 25oC. (a) N2(g) + O2(g) → 2NO(g) ΔG0 = kJ/mol (b) H2O(l) → H2O(g) ΔG0 = kJ/mol (c) 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l) ΔG0 = kJ/mol
Consider a general reaction A<-> B; Given that ΔG0= -RTlnK; (a) if ΔG0 <0, what can you predict about value of K? what can you say about the reaction? (b) if ΔG0 >0, what can you predict about value of K? what can you say about the reaction? (c ) if ΔG0 = +1.7kJ/mole, calculate the equilibrium constant K (assume T=298k)
what is the difference between ΔG° and ΔG"? Mainly, when can ljust use ΔGf to calculate ΔG°instead of using ΔG.-ΔΗ-TAS"? I hope this is clear.
What is the relationship between ΔG and the ΔG°F for the reaction below? MgF2(s) → Mg2+(aq) + 2 F-(aq) ΔG = {ΔG°f [Mg2+ (aq)] + 2 ΔG°f [F- (aq)] - ΔG°f [MgF2 (s)]} + RT ln ([Mg2+] [F-]2/[MgF2]) ΔG = {ΔG°f [Mg2+ (aq)] + 2 ΔG°f [F- (aq)]} + RT ln ([Mg2+] [F-]2) ΔG = {ΔG°f [Mg2+ (aq)] + 2 ΔG°f [F- (aq)] - ΔG°f [MgF2 (s)]} + RT ln ([Mg2+] [F-])2) ΔG = {ΔG°f [Mg2+ (aq)] + 2 ΔG°f...
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