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Find ΔS°rxn for the reaction of nitrogen monoxide with hydrogen to form ammonia and water vapor. Is the sign ofΔS°rxn as expected?

Respuesta :

The value of ΔS°rxn for the reaction is -310.9JK⁻¹ .

WHAT IS ΔS IN A REACTION ?

∆S is the change in entropy (disorder) from reactants to products. R is the gas constant (always positive) T is the absolute temperature (Kelvin, always positive) What it means: If ∆H is negative, this means that the reaction gives off heat from reactants to products.

General expression for ΔS°rxn of a chemical reaction is as follows:

[tex]\Delta S^o_{rxn} = \sum m S^o_{(products)} - \sum n S^o_{(reactants)}[/tex]

Here, m, n represent the stoichiometry coefficients of each product and reactant.

For the reaction,

2NO(g) + 5H₂(g) → 2NH₃(g) + 2H₂O(g)

The expression for ΔS°rxn is as follows:

[tex]\Delta S^o_{rxn} = [(2mol)(S^o_{NH_3 (g)}) + (2mol)(S^o_{H_2 O (g)})] [(2mol)(S^o_{NO (g)}) + (5mol)(S^o_{H_2 (g)})][/tex]

From the appendix B, the standard values for entropy S° for NH₃(g) is 193 J mol⁻¹ K⁻¹, for H₂O(g) is 188.72 J K⁻¹ mol⁻¹, for NO(g) the value of S° is 210.65 J mol⁻¹ K⁻¹, and S° for H₂(g) is 130.6 J mol⁻¹K⁻¹.

Substitute the standard values of entropy in the expression for ΔS°rxn.

[tex]\Delta S^o_{rxn} = [(2mol)(193Jmol^{-1}K^{-1}) + (2mol)(188.72Jmol^{-1}K^{-1})] -[/tex]

             [tex][(2mol)(210.65Jmol^{-1}K^{-1}) + (5mol)(130.6Jmol^{-1}K^{-1})][/tex]

          [tex]= [763.44 JK^{-1}] - [1074.3 JK^{-1}]\\= - 310.9 JK^{-1}[/tex]

Thus, the value of ΔS°rxn for this reaction is - 310.9 JK⁻¹.

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