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For the reaction MgSO3(s) → MgO(s) + SO2(g), which is spontaneous only at high temperatures, one would predict thata. ΔH˚ is negative and ΔS˚ is negativeb. ΔH˚ is positive and b. ΔS˚ is negativec. ΔH˚ is positive and ΔS˚ is positived. ΔH˚ is negative and ΔS˚ is positive

Respuesta :

Answer: c. ΔH˚ is positive and ΔS˚ is positive

Explanation:

According to Gibb's equation:

[tex]\Delta G=\Delta H-T\Delta S[/tex]

[tex]\Delta G[/tex] = Gibbs free energy  

[tex]\Delta H[/tex] = enthalpy change

[tex]\Delta S[/tex] = entropy change  

T = temperature in Kelvin

[tex]\Delta G[/tex]= +ve, reaction is non spontaneous

[tex]\Delta G[/tex]= -ve, reaction is spontaneous

[tex]\Delta G[/tex]= 0, reaction is in equilibrium

Thus for [tex]\Delta G=-ve[/tex]

Case :

[tex]T\Delta S[/tex] > [tex]\Delta H[/tex]

when [tex]\Delta S\text{ and }\Delta H[/tex] both have positive values.

[tex]\Delta G=(+ve)-T(+ve)[/tex]

[tex]\Delta G=(+ve)(-ve)=-ve[/tex]

Reaction is spontaneous only at at high temperatures.

Eduard22sly

Given that the reaction is spontaneous only at high temperature, one would predict that ΔH˚ is positive and ΔS˚ is positive (Option C)

Gibbs free energy

ΔG = ΔH – TΔS

Where

  • ΔG is the Gibbs free energy
  • ΔH is the enthalpy change
  • T is the temperature
  • ΔS is the change in entropy

NOTE

  • ΔG = +ve (non spontaneous)
  • ΔG = 0 (equilibrium)
  • ΔG = –ve (spontaneous)

Since the reaction is at high temperature, then TΔS must be greater than ΔH for ΔG to be negative (i.e spontaneous).

Thus, ΔH and ΔS must be positive for the reaction to be spontaneous as illustrated below:

  • ΔH = +ve
  • ΔS = +ve
  • ΔG =?

ΔG = ΔH – TΔS

ΔG = (+ve) – T(+ve)

ΔG = –ve

Learn more about Gibbs free energy:

https://brainly.com/question/9552459

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