for the reaction to generate 2‐phosphoglycerate (2PG) from 3‐phosphoglycerate (3PG, ΔG = 0.83 kJ/mol; ΔGo’ = 4.4 kJ/mol. (R = 8.31 x 10‐3 kJ•K‐1•mol‐1)

a)What is the ratio of 3PG:2PG in cells at 37C? I got 5.52 as answer for this part but i don't understand part b

equation I used for part a was: ΔGo’= -RT. K

b)Considering your result, is the reaction under cellular conditions favoring the formation of 3PG or
the formation of 2PG?

To answer part a, you correctly used the equation ΔGo' = -RT ln(K), where ΔGo' is the standard Gibbs free energy change, R is the gas constant (8.31 x 10-3 kJ•K-1•mol-1), T is the temperature in Kelvin, and K is the equilibrium constant. Rearranging the equation, we get K = e^(-ΔGo' / RT).

You are given ΔGo' = 4.4 kJ/mol and T = 37°C = 37 + 273.15 = 310.15 K. Plugging in these values into the equation, we have:

K = e^(-4.4 kJ/mol / (8.31 x 10-3 kJ•K-1•mol-1 × 310.15 K))
= e^(-0.005346 mol-1)
= 0.99466

Therefore, the ratio of 3PG to 2PG in cells at 37°C is approximately 1:0.99466, or 1.00534:1.

Now, to answer part b, we need to analyze whether the reaction under cellular conditions favors the formation of 3PG or the formation of 2PG. Since the ratio of 3PG to 2PG is greater than 1, this indicates that there is a higher concentration of 3PG compared to 2PG.

In a reaction at equilibrium, the reaction is equally likely to go in either direction. However, a higher concentration of reactants compared to products typically suggests that the reaction is favoring the formation of products.

Therefore, under cellular conditions, the reaction is favoring the formation of 3PG.