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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

Respuesta :

santoniaturner12
The answer to your question is -382kj I think

Answer: The [tex]\Delta G[/tex] for the reaction is -382 kJ.

Explanation:

For the following reaction:

[tex]2NH_3(g)+2O_2(g)\rightarrow NH_4NO_3(s)+H_2O(l)[/tex]

  • Equation used to calculate [tex]\Delta H_{rxn}[/tex] is:

[tex]\Delta H_{rxn}=\sum \Delta H_{products}-\sum \Delta H_{reactants}[/tex]

We are given:

[tex]\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[/tex]

[tex]\Delta H_{rxn}[/tex] for the reaction is calculated by:  

[tex]\Delta H_{rxn}=[1(\Delta H_{NH_4NO_3})+1(\Delta H_{H_2O})]-[2(\Delta H_{NH_3})+2(\Delta H_{O_2})][/tex]

Putting values in above equation, we get:

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

  • Equation used to calculate [tex]\Delta S_{rxn}[/tex] is:

[tex]\Delta S_{rxn}=\sum \Delta S_{products}-\sum \Delta S_{reactants}[/tex]

We are given:

[tex]\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[/tex]

[tex]\Delta S_{rxn}[/tex] for the reaction is calculated by:  

[tex]\Delta S_{rxn}=[1(\Delta S_{NH_4NO_3})+1(\Delta S_{H_2O})]-[2(\Delta S_{NH_3})+2(\Delta S_{O_2})][/tex]

Putting values in above equation, we get:

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

  • Now, to calculate [tex]\Delta G[/tex], the equation used is:

[tex]\Delta G=\Delta H-T\Delta S[/tex]

We are given:

[tex]\Delta H=-559170J\\T=298K\\\Delta S=-573.91J/K\\[/tex]

Putting values in above equation, we get:

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

Hence, the [tex]\Delta G[/tex] for the reaction is -382 kJ.

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