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  3. c. for this reaction in heart muscle, Δg°â

c. For this reaction in heart muscle, ΔG°’ = +4.7 kJ/mol but ΔG = -0.6 kJ/mol....

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Chemistry

c. For this reaction in heart muscle, ΔG°’ = +4.7 kJ/mol but ΔG= -0.6 kJ/mol. i. Explain, in words, how ΔG can be negative whenΔG°’ is positive. ii. What would be the ratio of 3PG to 2PG if thereaction were at equilibrium at 25°C? iii. What is the actual ratioof 3PG to 2PG in heart muscle (T = 37°C)?

Answer & Explanation Solved by verified expert
3.6 Ratings (669 Votes)

Given ΔG°’ = +4.7 kJ/mol = 4700 J/mol and ΔG = -0.6 kJ/mol = -600 J/mol

We know that ΔG = ΔG° - RT lnK

                          = ΔG°’ - RT lnK

                          = (+ve) - ( (+ve)(+ve) (+ve))

                          = +ve -(+ve)

                           = -ve

(ii) ΔG = ΔG° - RT lnK

Where R = gas constant = 8.314 J/(mol-K)

          T = Temperature = 25 oC = 25+273 = 298 K

Plug the values we get

600 = 4700 -(8.314x298 x ln K)

ln K = 1.65

   K = e1.65 = 5.23

    3PG <----> 2PG

But Equilibrium constant , K = [PG]2 / [PG]3 = 5.3

Simillarly the third one

                                   
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