Question

What is the ∆G for the following reaction under standard conditions (T = 298 K) for the formation of NH4NO3(s)? 2NH3(g) + 2O2(g) NH4NO3(s) + H2O(l) Given: NH4NO3(s): ∆Hf = -365.56 kJ ∆Sf = 151.08 J/K. NH3(g): ∆Hf = -46.11 kJ ∆Sf = 192.45 J/K. H2O(l): ∆Hf = -285.830 kJ ∆Sf = 69.91 J/K. O2(g): ∆Hf = 0.00 kJ ∆Sf = 205 J/K. 186.6 kJ 6.9 kJ -10.4 kJ -126.3 kJ -382 kJ

Answer 1

The answer to your question is -382kj I think

Answer 2

Answer: The
`\Delta G` for the reaction is -382 kJ.

Step-by-step explanation:

For the following reaction:

`2NH_3(g)+2O_2(g)\rightarrow NH_4NO_3(s)+H_2O(l)`

• Equation used to calculate
  `\Delta H_(rxn)` is:

`\Delta H_(rxn)=\sum \Delta H_(products)-\sum \Delta H_(reactants)`

We are given:

`\Delta H_(NH_3)=-46.11kJ/mol\\\Delta H_(O_2)=0.00kJ/mol\\\Delta H_(NH_4NO_3)=-365.56kJ/mol\\\Delta H_(H_2O)=-285.830kJ/mol`

`\Delta H_(rxn)` for the reaction is calculated by:

`\Delta H_(rxn)=[1(\Delta H_(NH_4NO_3))+1(\Delta H_(H_2O))]-[2(\Delta H_(NH_3))+2(\Delta H_(O_2))]`

Putting values in above equation, we get:

`\Delta H_(rxn)=[1(-365.56)+1(-285.83)]-[2(-46.11)+2(0)]kJ\\\\\Delta H_(rxn)=-559.17kJ=559170J`

• Equation used to calculate
  `\Delta S_(rxn)` is:

`\Delta S_(rxn)=\sum \Delta S_(products)-\sum \Delta S_(reactants)`

We are given:

`\Delta S_(NH_3)=192.45J/K\\\Delta S_(O_2)=205J/K\\\Delta S_(NH_4NO_3)=151.08J/K\\\Delta S_(H_2O)=69.91J/K`

`\Delta S_(rxn)` for the reaction is calculated by:

`\Delta S_(rxn)=[1(\Delta S_(NH_4NO_3))+1(\Delta S_(H_2O))]-[2(\Delta S_(NH_3))+2(\Delta S_(O_2))]`

Putting values in above equation, we get:

`\Delta S_(rxn)=[1(151.08)+1(69.91)]-[2(192.45)+2(205)]J/K\\\\\Delta S_(rxn)=-573.91J/K`

• Now, to calculate
  `\Delta G`, the equation used is:

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

We are given:

`\Delta H=-559170J\\T=298K\\\Delta S=-573.91J/K\\`

Putting values in above equation, we get:

`\Delta G=(-559170J)-[298K* (-573.91J/K)]\\\\\Delta G=-382kJ`

Hence, the
`\Delta G` for the reaction is -382 kJ.