Respuesta :
When adjusted for any changes in δh and δs with temperature, the standard free energy change δg∘t at 2400 k is equal to 1.22×105j/mol, then the equilibrium constant at 2400 k is 2.21×10−3. The answer to the statement is 2.21×10−3.
The equilibrium constant of a reaction is related the Gibbs free energy in a natural logarithm relationship. It is expressed as:
G = - RT ln K
where R is 8.314 and T is the temperature. From this, we can easily calculate for the change in the equilibrium constant at different temperature.
1.22×105 = - 8.314 (2400) ln K
K = 2.21x10^-3
G = - RT ln K
where R is 8.314 and T is the temperature. From this, we can easily calculate for the change in the equilibrium constant at different temperature.
1.22×105 = - 8.314 (2400) ln K
K = 2.21x10^-3