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

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

Answer 2

Answer:

$\Delta G^o=-5.4032 kJ$

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

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 $\Delta G^o=0$:

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

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

$T=\frac{\Delta H^o}{\Delta S^o}$

$T=\frac{-58.03 kJ}{-0.1766 kJ/K}$

$T=328.6 K$