Respuesta :
Answer:
The correct answer is option b.
Explanation:
[tex]SO_2 (g)+NO_2 (g)\rightarrow SO_3 (g)+NO (g)[/tex]
The equation used to calculate Gibbs free change is of a reaction is:
[tex]\Delta G^o_{rxn}=\sum [\Delta G^o_f(product)]-\sum [\Delta G^o_f(reactant)][/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta G^o_{rxn}=(\Delta G^o_f_{(SO_3(g))}+\Delta G^o_f_{(NO(g))})-(\Delta G^o_f_{(SO_2(s))}+G^o_f_{(NO_2(g))})[/tex]
We are given:
[tex]\Delta G^o_f_{(SO_2(s))}=-300.2 kJ/mol\\\Delta G^o_f_{(NO_2(g))}=51 kJ/mol[/tex]
[tex]\Delta G^o_f_{(SO_3(s))}=-371 kJ/mol\\\Delta G^o_f_{(NO(g))}=86.6kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta G^o_{rxn}=(-371 kJ/mol+86.6kJ/mol)-(-300.2 kJ/mol+51 kJ/mol)=-35.2 kJ/mol[/tex]
To calculate the [tex]K_c[/tex] (at 298 K) for given value of Gibbs free energy, we use the relation:
[tex]\Delta G^o=-RT\ln K_c[/tex]
where,
[tex]\Delta G^o[/tex] = Gibbs free energy = -35.2 kJ/mol = -35200 J/mol (Conversion factor: 1kJ = 1000J)
R = Gas constant = [tex]8.314J/K mol[/tex]
T =Temperature =298K[/tex]
[tex]K_1[/tex] = equilibrium constant at 298 k;
Putting values in above equation, we get:
[tex]-35200 J/mol=-(8.314J/Kmol)\times 298K\times \ln K_c\\\\K_c=1.479\times 10^6\approx 1.5\times 10^6[/tex]
Hence, correct answer is option b.
