Answer : At temperatures greater than [tex]755.9^oC[/tex] this reaction is spontaneous under standard conditions.
Explanation : Given,
[tex]\Delta H[/tex] = 131.3 KJ/mole = 131300 J/mole
[tex]\Delta S[/tex] = 127.6 J/mole.K
Gibbs–Helmholtz equation is :
[tex]\Delta G=\Delta H-T\Delta S[/tex]
As per question the reaction is spontaneous that means the value of [tex]\Delta G[/tex] is negative or we can say that the value of [tex]\Delta G[/tex] is less than zero.
[tex]\Delta G<0[/tex]
The above expression will be:
[tex]0>\Delta H-T\Delta S[/tex]
[tex]T\Delta S>\Delta H[/tex]
[tex]T>\frac{\Delta H}{\Delta S}[/tex]
Now put all the given values in this expression, we get :
[tex]T>\frac{131300J/mole}{127.6J/mole.K}[/tex]
[tex]T>1028.99K[/tex]
[tex]T>755.9^oC[/tex]
Therefore, at temperatures greater than [tex]755.9^oC[/tex] this reaction is spontaneous under standard conditions.
The complete statement will be as follows . At temperatures greater than T>755.9^C ∘C this reaction is spontaneous under standard conditions.
Question Parameter(s):
C(s)+H2O(g)→CO2(g)+H2(g)
ΔH∘=131.3kJ/mol
ΔS∘=127.6J/K⋅mol at 298K.
Generally, the equation for the reaction is mathematically given as
C(s)+H2O(g)→CO2(g)+H2(g)
Where
dG=dH-T dS
[tex]T > \frac{d H}{d S}\\\\T > \frac{131300}{127.6}[/tex]
T>755.9^C
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