3 years ago

Yo fm on ig chs_luucy. ill give u brainleist .

Chemistry

2 answers:

Alex3 years ago

8 0

Answer:

this is a homework app, not a tell people to follow your ig app

Explanation:

xz_007 [3.2K]3 years ago

7 0

Alright bet! Say less

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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 v

Ierofanga [76]

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 $\Delta G^o=0$ :

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

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

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

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

$T=328.6 K$

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3 years ago

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