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Use the table of standard entropy values to determine º for the following reaction: 2CH3OH(l) + 3O2(g) → 2CO2(g) + 4 H2O(g)

+70.627 J/mol·K



– 70.627 J/mol·K



+313.766 J/mol·K



–313.766 J/mol·K

Respuesta :

Standar molar entropies from a table:

CH3OH (l): 126.8 J / K*mol
O2 (g): 205.1 J/ K*mol
CO2(g): 213.7 J/K*mol
H2O(g): 188.8 J/J*mol

Now use the coefficients of the reaction and sum the product entropies less the reactant entropies:

4*188.8 + 2*213.7 - 3*205.1 - 2* 126.8 = 313.7 J/mol*K

Answer: option C) +3173.766 J/mol*K 
lhmarianateixeira

The entropy value of the products of this reaction will be +3173.766 J/mol*K

what is entropy?

Entropy is an important physical quantity used in Statistical Mechanics and Thermodynamics to measure the degree of disorder in a system. We say that the greater the entropy change of a system, the greater its disorder, that is, the less energy will be available to be used.

Standar molar entropies from a table:

  • CH3OH (l): 126.8 J / K*mol
  • O2 (g): 205.1 J/ K*mol
  • CO2(g): 213.7 J/K*mol
  • H2O(g): 188.8 J/J*mol

Using the equation given in the statement, we have:

[tex]2CH3OH(l) + 3O2(g) \rightarrow 2CO2(g) + 4 H2O(g)[/tex]

Now use the coefficients of the reaction and sum the product entropies less the reactant entropies:

[tex]4*188.8 + 2*213.7 - 3*205.1 - 2* 126.8 = 313.7 J/mol*K[/tex]

See more about entropy at brainly.com/question/13135498

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