Question

Given the data:

N2(g), ?H°f= 0.00 kJ mol-1, S° = +191.5 J mol-1 K-1
H2(g), ?H°f = 0.00 kJ mol-1, S° = +130.6 J mol-1 K-1
NH3(g), ?H°f= ?46.0 kJ mol-1, S° = +192.5 J mol-1 K-1
calculate the standard free energy change, ?G° for the reaction:
N2(g) + 3 H2(g) --> 2 NH3(g)

A. +112.3 kJ
B. ?7.4 kJ
C. ?32.9 kJ
D. ?84.6 kJ

Answer 1

Answer: The standard Gibbs free energy change of the reaction is -32.9 kJ

Step-by-step explanation:

For the given chemical equation:

`N_2(g)+3H_2(g)\rightarrow 2NH_3(g)`

• The equation for the enthalpy change of the above reaction is:

`\Delta H^o_(rxn)=[(2* \Delta H^o_f_((NH_3(g))))]-[(1* \Delta H^o_f_((N_2(g))))+(3* \Delta H^o_f_((H_2(g))))]`

We are given:

`\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`

Putting values in above equation, we get:

`\Delta H^o_(rxn)=[(2* (-46.0))]-[(1* 0)+(3* 0)]=-92kJ=-92000J`

• The equation for the entropy change of the above reaction is:

`\Delta S^o_(rxn)=[(2* \Delta S^o_((NH_3(g))))]-[(1* \Delta S^o_((N_2(g))))+(3* \Delta S^o_((H_2(g))))]`

We are given:

`\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`

Putting values in above equation, we get:

`\Delta S^o_(rxn)=[(2* (192.5))]-[(1* (191.5))+(3* (130.6))]=-198.3J/K`

• To calculate the standard Gibbs free energy of the reaction, we use the equation:

`\Delta G^o=\Delta H^o-T\Delta S^o`

where,

`\Delta H^o` = standard enthalpy change = -92000 J

T = Temperature =
`25^oC=[273+25]K=298K`

`\Delta S^o` = standard entropy of the reaction = -198 J/K

Putting values in above equation, we get:

`\Delta G^o=-92000J-(298K* (-198J/K))\\\\\Delta G^o=-32996J=-32.9kJ`

Hence, the standard Gibbs free energy change of the reaction is -32.9 kJ