Two biochemical reactions have the same Keq 5 x 108 at temperature T1 298 K However, Reaction 1 has H 28 kJ mol 1 and Reaction 2 has H 28 kJmol 1 The two reactions utilize the same reactants Your lab partner has proposed that you can get more of the reactants to proceed via Reaction 2 rather than R...

This strategy will NOT work because Reaction 1 has a negative enthalpy c ...

Two biochemical reactions have the same Keq = 5 x 108 at temperature T1= 298 K. However,

Question:

Two biochemical reactions have the same Keq = 5 x 108 at temperature T1= 298 K. However, Reaction 1 has ΔH° = -28 kJ ∙ mol-1 and Reaction 2 has ΔH° =-28 kJ∙mol-1. The two reactions utilize the same reactants. Your lab partner has proposed that you can get more of the reactants to proceed via Reaction 2 rather than Reaction 1 by lowering the temperature of the reaction. Will this strategy work? Why or why not? How much would the temperature have to be raised or lowered in order to change the value of K2 / K1 from 1 to 10?