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Consider the following reaction at 298 K:

2H2(g) + O2(g) --> 2H2O(g) DeltaH= -483.6 kj

Calculate the following quantities.

Delta S sys=____________
Delta S surr=__________
Delta S univ=___________

Respuesta :

Answer :

[tex]\Delta S_{sys}[/tex] = -1622.8 J/K

[tex]\Delta S_{surr}[/tex] = -94.6 J/K

[tex]\Delta S_{univ}[/tex] = 0 J/K

Explanation :  Given,

[tex]\Delta H[/tex] = -483.6 kJ

The given chemical reaction is:

[tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(g)[/tex]

First we have to calculate the value of [tex]\Delta S_{sys}[/tex].

[tex]\Delta S_{sys}=\frac{\Delta H}{T}[/tex]

[tex]\Delta S_{sys}=\frac{-483.6kJ}{298K}[/tex]

[tex]\Delta S_{sys}=-1.6228kJ/K=-1622.8J/K[/tex]

Entropy of system = -1622.8 J/K

As we know that:

Entropy of system = -Entropy of surrounding = 1622.8 J/K

and,

Entropy of universe = Entropy of system + Entropy of surrounding

Entropy of universe = -1622.8 J/K + (1622.8 J/K)

Entropy of universe = 0

Eduard22sly

Given the data from the question, the entropy of system, surrounding and universe are

A. The entropy of the system is –1622.8 J/K

B. The entropy of the surrounding is 1622.8 J/K

C. The entropy of the universe is 0 J/K

A. How to determine the entropy of the system

From the question given above, the following data were obtained:

  • Enthalpy change (ΔH) = –483.6 KJ
  • Temperature (T) = 298 K
  • Change in entropy of system (ΔSₛᵧₛ) =?

ΔS = ΔH / T

ΔSₛᵧₛ = –483.6 / 298

ΔSₛᵧₛ = –1.6228

ΔSₛᵧₛ = –1.6228 × 1000

ΔSₛᵧₛ = –1622.8 J/K

B. How to determine the entropy of the surrounding

  • Change in entropy of system (ΔSₛᵧₛ) = –1622.8 J/K
  • Change in entropy of surrounding (ΔSₛᵤᵣ) =?

ΔSsys = –ΔSsur

–1622.8 = –ΔSsur

ΔSₛᵤᵣ = 1622.8 J/K

C. How to determine the entropy of the universe

  • Change in entropy of system (ΔSₛᵧₛ) = –1622.8 J/K
  • Change in entropy of surrounding (ΔSₛᵤᵣ) = 1622.8 J/K
  • Change in entropy of universe (ΔSᵤₙᵢ) =?

ΔSᵤₙᵢ = ΔSₛᵧₛ + ΔSₛᵤᵣ

ΔSᵤₙᵢ = –1622 + 1622

ΔSᵤₙᵢ = 0 J/K

Learn more about entropy:

https://brainly.com/question/13834300

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