Question

For a certain chemical reaction, the standard Gibbs free energy of reaction at 10.0 °C is 149. kJ. Calculate the equilibrium constant K for this reaction to 2 significant digits. Round your answer K ?

Answer 1

Answer:

This is confusing

Explanation:

Answer 2

Answer: The equilibrium constant for this reaction is $3.1\times 10^{-28}$

Explanation:

To calculate the equilibrium constant (at 15°C) for given value of Gibbs free energy, we use the relation:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = 149. kJ/mol = 149000 J/mol  (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = $8.314J/K mol$

T = temperature = $15^oC=[273+15]K=283K$

$K_{eq}$ = equilibrium constant at 10°C = ?

Putting values in above equation, we get:

$149000J/mol=-(8.314J/Kmol)\times 283K\times \ln K_{eq}\\\\K_{eq}=3.1\times 10^{-28}$

Hence, the equilibrium constant for this reaction is $3.1\times 10^{-28}$