For the reaction at 298 K, 2NO2(g) N2O4(g) the values of ΔH° and ΔS° are -58.03 kJ and -176.6 J/K, respectively. Calculate the value of ΔG° at 298 K. ΔG° = kJ Assuming that ΔH° and ΔS° do not depend on temperature, at what temperature is ΔG° = 0? T = K

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Question:

For the reaction at 298 K, 2NO2(g) N2O4(g) the values of ΔH° and ΔS° are -58.03 kJ and -176.6 J/K, respectively. Calculate the value of ΔG° at 298 K. ΔG° = kJ Assuming that ΔH° and ΔS° do not depend on temperature, at what temperature is ΔG° = 0? T = K

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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 KSolving:[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 kJT = 328.6 K Explanation:DataΔH°: -58.03 kJΔS: -176.6 J/K = -0.1766 kJ/KThe change in Gibbs free energy is defined as:ΔG° = ΔH° − T*ΔS°When T = 298 K:ΔG° = -58.03 − 298*(-0.1766) = -5.4032 kJif ΔG° = 0 kJ, then:0 = -58.03 − T*(-0.1766)58.03 = T*0.1766T = 58.03/0.1766T = 328.6 K

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