Respuesta :
Answer: The equilibrium constant for this reaction is [tex]3.1\times 10^{-28}[/tex]
Explanation:
To calculate the equilibrium constant (at 15°C) for given value of Gibbs free energy, we use the relation:
[tex]\Delta G^o=-RT\ln K_{eq}[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs free energy = 149. kJ/mol = 149000 J/mol (Conversion factor: 1 kJ = 1000 J )
R = Gas constant = [tex]8.314J/K mol[/tex]
T = temperature = [tex]15^oC=[273+15]K=283K[/tex]
[tex]K_{eq}[/tex] = equilibrium constant at 10°C = ?
Putting values in above equation, we get:
[tex]149000J/mol=-(8.314J/Kmol)\times 283K\times \ln K_{eq}\\\\K_{eq}=3.1\times 10^{-28}[/tex]
Hence, the equilibrium constant for this reaction is [tex]3.1\times 10^{-28}[/tex]
