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

Answer 1

ΔG° = -5.4032 kJ

T = 328.6 K

Step-by-step 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

Answer 2

`\Delta G^o=-5.4032 kJ`

The temperature for
`\Delta G^o=0[/tex is [tex]T=328.6 K`

Step-by-step explanation:

The three thermodinamic properties (enthalpy, entropy and Gibbs's energy) are linked in the following formula:

`\Delta G^o=\Delta H^o + T*\Delta S^o`

Where:

`\Delta G^o` is Gibbs's energy in kJ

`\Delta H^o` is the enthalpy in kJ

`\Delta S^o` is the entropy in kJ/K

`T` is the temperature in K

Solving:

`\Delta G^o=-58.03 kJ - 298K*-0.1766 kJ/K`

`\Delta G^o=-5.4032 kJ`

For
`\Delta G^o=0`:

`0=\Delta H^o - T*\Delta S^o`

`\Delta H^o= T*\Delta S^o`

`T=(\Delta H^o)/(\Delta S^o)`

`T=(-58.03 kJ)/(-0.1766 kJ/K)`

`T=328.6 K`