Question

The value of delta G for the phosphorylation of glucose in glycolysis is 13.8 kj/mol. Find the value of Kc at 298K?

Answer 1

ΔG=-2.303RT log Kc
where R =8.314jk^-1mol^-1;T=298k;
Given
ΔG=13.8KJ/Mol but has value of R is in joules,we take it as 13.8×10³j/mol
therefore 13.8×10³=-2.303×8.314×298logKc
log Kc =-2.4185
antilog=0.2661×10^-2

Answer 2

Answer: The value of $K_c$ at 298 K is $3.81\times 10^{-3}$

Explanation:

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_c$

where,

$\Delta G^o$ = standard Gibbs free energy = 13.8 kJ/mol = 13800 J/mol  (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = $8.3145J/K mol$

T = temperature = 298 K

$K_c$ = equilibrium constant in terms of concentration = ?

Putting values in above equation, we get:

$13800J/mol=-(8.3145J/Kmol)\times 298K\times \ln (K_c)\\\\K_c=e^{-5.57}=3.81\times 10^{-3}$

Hence, the value of $K_c$ at 298 K is $3.81\times 10^{-3}$