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Given the standard entropy values in the table, what is the value of `Delta"Sº"_"reaction"` for the following reaction?

2NiS(s) + 3O2(g) → 2SO2(g) + 2NiO(s)
A. –149 J/mol·K
B. +149 J/mol·K
C. +28 J/mol·K
D. – 28 J/mol·K

Given the standard entropy values in the table what is the value of DeltaSºreaction for the following reaction 2NiSs 3O2g 2SO2g 2NiOs A 149 JmolK B 149 JmolK C class=

Respuesta :

Answer:

entropy of reaction = 572-718= -146J/mol-k.

Explanation:

entropy of a reaction = entropy of product - entropy of reactant.

entropy of product = 2×entropy of SO2+2×entropy of NiO

                               =[tex]2\times 248+2\times 38[/tex]

                              =572 J/mol-k

entropy of reactant = 2×entropy if NiS+3×entropy of O2

                                =[tex]2\times 53+3\times 205[/tex]

                                =718 J/mol-k

therefore entropy of reaction = 572-718= -146J/mol-k.

sebassandin

Answer:

[tex]\Delta S^o=-149\frac{J}{mol*K}[/tex]

Explanation:

Hello,

In this case, we define the standard change in the entropy for a chemical reaction as:

[tex]\Delta S^o=\Sigma \nu _iS_i^o[/tex]

Whereas [tex]\nu _i[/tex] is the stoichiometric coefficient of the species i which is negative for products and positive for reactants and [tex]S_i^o[/tex] is the standard entropy for the species i in J/mol*K. Thus, we have:

[tex]\Delta S^o=(-2)(53} )+(-3)(205)+(2)(248)+(2)(38)\\\Delta S^o=-149\frac{J}{mol*K}[/tex]

Best regards.

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