Respuesta :
Answer:
The equilibrium constant for this reaction at 298.15 K is [tex]2.067\times 10^{13}[/tex].
Explanation:
The equation used to calculate Gibbs free change is of a reaction is:
[tex]\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f(product)]-\sum [n\times \Delta G^o_f(reactant)][/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta G^o_{rxn}=(2\times \Delta G^o_f_{(H_2O(g))}+2\times G^o_f_{(Cl_2(g))})-(4\times \Delta G^o_f_{(HCl(g))}+1\times G^o_f_{(O_2(g))})[/tex]
We are given:
[tex]\Delta G^o_f_{(HCl(g))}=-95.3 kJ/mol\\\Delta G^o_f_{(H_2O(g))}=228.6 kJ/mol[/tex]
[tex]\Delta G^o_f_{(O_2(g))}=\Delta G^o_f_{(Cl_2(g))}=0[/tex] (pure element)
Putting values in above equation, we get:
[tex]\Delta G^o_{rxn}=(2\times (-228.6 kJ/mol)+2\times 0 kJ/mol)-(4\times -95.3 kJ/mol+1\times 0 kJ/mol)=-76 kJ/mol[/tex]
To calculate the [tex]K_1[/tex] (at 25°C) for given value of Gibbs free energy, we use the relation:
[tex]\Delta G^o=-RT\ln K_1[/tex]
where,
[tex]\Delta G^o[/tex] = Gibbs free energy = -76 kJ/mol = -76000 J/mol
(Conversion factor: 1kJ = 1000J)
R = Gas constant = [tex]8.314J/K mol[/tex]
T = temperature = 298.15 K[/tex]
[tex]K_1[/tex] = equilibrium constant at 25°C = ?
[tex]-76000 J/mol=-8.314J/K mol\times 298.145 K\ln K_1[/tex]
[tex]\ln K_1=\frac{-76000 J/mol}{-8.314J/K mol\times 298.15 K}[/tex]
[tex]K_1=2.067\times 10^{13}[/tex]
The equilibrium constant for this reaction at 298.15 K is [tex]2.067\times 10^{13}[/tex].
