Question

The free energy (AG) for a reaction under a set of initial conditions/concentrations is -14.3 kcal/mol. The standard free energy (AG°) of this reaction is -6.0 kcal/mol. Given this information, what will be the value of the ratio of the reaction quotient (Q) to the equilibrium constant (Kea)? In other words, what will Q/Keg equal? Assume a temperature of 25°C. The value of the gas constant is 1.987 × 10-3 kcal/K-mol. To make your answer feasible to enter in Canvas, take the log (base 10) of your Q/Keq value and enter this result to the nearest tenth

Answer 1

Sol . As deltaG° = - 2.303 × R × T × log(Keq)  

where deltaG° = standard free energy of reaction = -6.0 Kcal / mol   

R = gas constant = 1.987 × 10^-3 Kcal / K mol  

T = Temperature = 25°C = 298 K

Keq = Equilibrium constant

So , log(Keq) = - deltaG° / (2.303 × R × T )

= - ( -6.0) / ( 2.303 × 1.987 × 10^-3 × 298 )  

= 4.3999

and , Keq = 10^4.3999 = 25113.081

Now , deltaG = deltaG° - 2.303 × R × T × log(Q)

where deltaG = free energy of reaction = - 14.3 Kcal / mol

Q = Reaction quotient

So , - 14.3 = - 6.0 - 2.303 × 1.987 × 10^-3 × 298 × log(Q)

- 8.3 = - 2.303 ×  1.987 × 10^-3 × 298 × log(Q)

So , log(Q) = 6.0865

and , Q = 10^6.0865 = 1220393.821

Therefore , Q / Keq = 1220393.821 / 25113.081  

= 48.596