Question

like me to grade. I vow did the take homeow 1. In glycolysis, the enzyme pyruvate the enzyme pyruvate kinase catalyzes this reaction phosphoenolpyruvate + ADP - pyruvate + ATP caction using the information provided a. What is the equilibrium constant for below? (R-8.315 J/mol K; T = 298 K) Reaction 1: ATP - ADP +P Reaction 2: phosphoenolpyruvate vale pyruvate + PAG AG"=-30.5 kJ/mol pyruvate + P AGⓇ = -61.9 kJ/mol ntrations are provided below, what is the actual b. If the reactant and product concent free energy change? (ATP)=1.0 mm; (ADP) - 0.1 mm; [P]- 1.0 mM: (PEP)=0.023 mM: LP M. (PEP) = 0,023 mM: [pyruvate]=0.051 mm s6=06"+RT In Keg. Inkeq = (-61.9 kJ/ma--30-SKJ/mol). DG" = RT ln keq Ilmol --30-SKJ/mol) +1-0083 kJ/molk) (298k) Inkeq = -92.4 kJ/mol +2.47 kJ/mol Inkeq = -89.93 kJ/mol e -89.93=8.810-40 G = -30500J/molkt 8 318 Smor (298k) in tam 1 = -3026383 Llomm , here use AG=AGO RTG combined Concentration of overall products/ eachta AG = -619005/molkt 8.315 5lmol (298k) in/ 1.051 mm "L0.823mm ) = -52,409.57 = -5o una Timol Page 6

Answer 1

Solution:

Part A)

The standard free energy change is obtained by inverting reaction 1 and then adding in reaction 2.

Thus,

ΔG° = -61.9 kJ/mol - (-30.5) kJ/mol = -31.4 kJ/mol

Since, standard free energy change (ΔG°) is related with equilibrium constant (keq) as,

ΔG° = - RT ln Keq

ln keq = - ΔG°/RT

lnkeq = - (-31.4) kJ mol-1 / 8.315 J k mol-1 x 298 K

ln keq = 31400 J mol-1 / 2477.87 J mol-1

ln keq = 12.67

log keq = 12.67 / 2.303

(since , ln = 2.303 log)

log keq = 5.50

keq = antilog 5.50 = 3.16 x 10^5

keq = 3.16 x 10^5

Part B)

ΔG = ΔG° + RT ln Q

= ΔG° + RT ln [Pyruvate] [ATP] / [ADP] [PEP]

= -31.4 kJ mol-1 + RT ln (0.051mM) (1mM) / (0.1mM) (0.023mM)

= - 31.4 kJ mol-1 + 8.315 J K-1 mol-1 x 298 K x ln 0.051 / 0.0023

= - 31.4 kJ mol-1 + 2.48 kJ mol-1 x ln 22.17

= -31.4 kJ mol-1 + 2.48 x 3.10 kJ mol-1

= -31.4 kJ mol-1 + 7.7 kJ mol-1

ΔG = -23.7 kJ mol-1