Answer:
[tex]\Delta G^o=-5.4032 kJ[/tex]
The temperature for [tex]\Delta G^o=0[/tex is [tex]T=328.6 K[/tex]
Explanation:
The three thermodinamic properties (enthalpy, entropy and Gibbs's energy) are linked in the following formula:
[tex]\Delta G^o=\Delta H^o + T*\Delta S^o[/tex]
Where:
[tex]\Delta G^o[/tex] is Gibbs's energy in kJ
[tex]\Delta H^o[/tex] is the enthalpy in kJ
[tex]\Delta S^o[/tex] is the entropy in kJ/K
[tex]T[/tex] is the temperature in K
Solving:
[tex]\Delta G^o=-58.03 kJ - 298K*-0.1766 kJ/K[/tex]
[tex]\Delta G^o=-5.4032 kJ[/tex]
For [tex]\Delta G^o=0[/tex]:
[tex]0=\Delta H^o - T*\Delta S^o[/tex]
[tex]\Delta H^o= T*\Delta S^o[/tex]
[tex]T=\frac{\Delta H^o}{\Delta S^o}[/tex]
[tex]T=\frac{-58.03 kJ}{-0.1766 kJ/K}[/tex]
[tex]T=328.6 K[/tex]
Answer:
ΔG° = -5.4032 kJ
T = 328.6 K
Explanation:
Data
ΔH°: -58.03 kJ
ΔS: -176.6 J/K = -0.1766 kJ/K
The change in Gibbs free energy is defined as:
ΔG° = ΔH° − T*ΔS°
When T = 298 K:
ΔG° = -58.03 − 298*(-0.1766) = -5.4032 kJ
if ΔG° = 0 kJ, then:
0 = -58.03 − T*(-0.1766)
58.03 = T*0.1766
T = 58.03/0.1766
T = 328.6 K
