Question

Consider a general reaction A(aq) = B(aq) The AG of the reaction is -3.490 kJ - mol. Calculate the equilibrium constant for the reaction at 25°C. Kin= 0.333 What is AG for the reaction at body temperature (37.0 °C) if the concentration of Ais 1.8 M and the concentration of Bis 0.65 M? AG -9.561 kJ mol

Answer 1

① Alaq) enzyme, B (ay) & Keg - Cone of product – [B7 n Come of reactant - AG of the site is given - – 3.490 kJ mot! = - 3.490 x 163 s moth Tem Þ(T) = 25°C = 25+273 - 29816. As we know AG= DG + RT enk) At equilibrium AG=0 & ka keq O2 AG + RT en keq. DG ₂ - RT en keq where R= 8.314 J K-1 mort In Req = - SG substituting all values dan keg - (-3.490 x103) 8.314x298 2.303 log keg - 3.490 x 103 8.314x296 log keg= 3.490X108 6:314x298X2-303 keg= - 0.2131

4 1 from egen keq = [B] = 0.65 -0.36 [A] 1.8 Temp (T)= 37C = 37+273=310k R = 8.314 J K mot Ay=- RT enkeg DG = - 8.314x310x0.36 06° -927.8.424 Imot! lor [AG®--019278424 ksmo!)