Respuesta :
Answer: The standard Gibbs free energy change of the reaction is -32.9 kJ
Explanation:
For the given chemical equation:
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]
- The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(NH_3(g))})]-[(1\times \Delta H^o_f_{(N_2(g))})+(3\times \Delta H^o_f_{(H_2(g))})][/tex]
We are given:
[tex]\Delta H^o_f_{(N_2(g))}=0.00kJ/mol\\\Delta H^o_f_{(H_2(g))}=0.00kJ/mol\\\Delta H^o_f_{(NH_3(g))}=-46.0kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(2\times (-46.0))]-[(1\times 0)+(3\times 0)]=-92kJ=-92000J[/tex]
- The equation for the entropy change of the above reaction is:
[tex]\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NH_3(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(3\times \Delta S^o_{(H_2(g))})][/tex]
We are given:
[tex]\Delta S^o_{(N_2(g))}=+191.5J/mol.K\\\Delta S^o_{(H_2(g))}=+130.6J/mol.K\\\Delta S^o_{(NH_3(g))}=+192.5J/mol.K[/tex]
Putting values in above equation, we get:
[tex]\Delta S^o_{rxn}=[(2\times (192.5))]-[(1\times (191.5))+(3\times (130.6))]=-198.3J/K[/tex]
- To calculate the standard Gibbs free energy of the reaction, we use the equation:
[tex]\Delta G^o=\Delta H^o-T\Delta S^o[/tex]
where,
[tex]\Delta H^o[/tex] = standard enthalpy change = -92000 J
T = Temperature = [tex]25^oC=[273+25]K=298K[/tex]
[tex]\Delta S^o[/tex] = standard entropy of the reaction = -198 J/K
Putting values in above equation, we get:
[tex]\Delta G^o=-92000J-(298K\times (-198J/K))\\\\\Delta G^o=-32996J=-32.9kJ[/tex]
Hence, the standard Gibbs free energy change of the reaction is -32.9 kJ
