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

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 when ΔG°’ is positive. ii. What would be the ratio of 3PG to 2PG if the reaction were at equilibrium at 25°C? iii. What is the actual ratio of 3PG to 2PG in heart muscle (T = 37°C)?

Solutions

Expert Solution

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 = e^1.65 = 5.23

3PG <----> 2PG

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

Simillarly the third one

queen_honey_blossom answered 1 year ago

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