1. If the G o ’ value for an isomerization reaction, A A’ is 2.7 kcal/mole...

1. If the G o ’ value for an isomerization reaction, A A’ is 2.7 kcal/mole at 37 C,

a) what is the ratio of A’ and A at equilibrium?

b) What is G’ for the reaction if A’ is reacted soon after it is formed and its concentration is reduced to 1/10 x the normal equilibrium concentration? The concentration of A doesn’t change and just consider the concentration effect on G’, not the linkage to a second reaction.

The box next to G is a delta symbol.

Solutions

Expert Solution

a)
delta G = 2.7 KCal/mol
                  = 2700 Cal/mol
                   = 2700*4.184 J/mol
                   = 11296.8 J/mol

equilibrium constant, Kc= [A'] / [A]
T = 37 oC = (273 + 37) K = 310 K

Use:
delta G = -R*T*ln Kc
11296.8 = -8.314* 310 * ln ( [A'] / [A])
ln ( [A'] / [A]) = -4.383
[A'] / [A] = 0.0125
Answer: 0.0125

b)
new Kc = [new A'] / [new A]
     = 1/10 * [A'] / [A] since new A' = 1/10 * A'
    = 1/10 * old Kc
     = 1/10* 0.0125
     = 0.00125

delta G'= -R*T*ln Kc
   = -8.314*310 * ln (0.00125)
   = 17228.5 J/mol
    = 17228.5 / 4.184 cal/mol
    = 4117.7 cal/mol
    = 4.12 KCal/mol
Answer: 4.12 KCal/mol

queen_honey_blossom answered 1 year ago

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