The standard free energy ( Δ G ∘ ′ ) (ΔG∘′) of the creatine kinase reaction is − 12.6 kJ ⋅ mol − 1 . −12.6 kJ⋅mol−1. The Δ G ΔG value of an in vitro creatine kinase reaction is − 0.1 kJ ⋅ mol − 1 . −0.1 kJ⋅mol−1. At the start of the reaction, the concentration of ATP is 5 mM, 5 mM, the concentration of creatine is 17 mM, 17 mM, and the concentration of creatine phosphate is 25 mM. 25 mM. Using the values given, calculate the starting concentration of ADP in micromolar.

The concentration of [ADP] = 21.896*10^-6 μM

Explanation:

Given Data:

creatine + ATP > ADP + creatine phosphate    

ΔG∘   = -12.6 KJ/mole  = -12600 J/mole

ΔG = -0.1 KJ/mole  =  -100 J/mole

[Creatine phosphate]  = 25 mM = 25*10^-3 M

[Creatine] = 17 mM    = 17*10^-3 M

[ATP]   =5mM = 5*10^-3M

Calculating the concentration of [ADP] using the formula;

ΔG = ΔG∘ + RTlnQc

Substituting, we have

-12600   = -100 + 8.314*298lnQc

-12600+100 = 8.314*298lnQc

-12500   = 2477.57lnQc

lnQc = -12500/2477.57

lnQc = -5.045

Qc = e^ -5.045

Qc   = 6.44*10^-3

But,

Qc    = [Creatine phosphate]*[ADP]/[creatine]*[ATP]

6.44*10^-3   = 25*10^-3*[ADP]/ (17*10^-3* 5*10^-3)

6.44*10^-3 = 25*10^-3[ADP]/8.5*10^-5

6.44*10^-3 * 8.5*10^-5 = 25*10^-3[ADP]

5.474*10^-7 = 25*10^-3[ADP]

[ADP] = 5.474*10^-7 /25*10^-3

          = 2.1896 *10^-5 M

          = 21.896*10^-6 μM

Therefore, the concentration of [ADP] = 21.896*10^-6 μM

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of c is as shown here in the image.