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.

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:

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

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