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The following reaction has the thermodynamic values at 298 K: ΔH° = -136.9 kJ/mol and ΔS° = -120.6 J/mol K. Calculate ΔG° at 298 K for this reaction in kJ/mol (Enter your answer to four significant figures.):

Respuesta :

Answer:

[tex]\Delta G^{0}[/tex] at 298 K is -101.0 kJ/mol

Explanation:

According to thermodynamic of state, [tex]\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}[/tex]

where, [tex]\Delta G^{0}[/tex], [tex]\Delta H^{0}[/tex], T and [tex]\Delta S^{0}[/tex] represent change in free energy in standard state, change in enthalpy in standard state, temperature in kelvin scale and change in entropy in standard state.

Here, [tex]\Delta H^{0}=-136.9kJ/mol[/tex], T = 298 K and [tex]\Delta S^{0}=-0.1206kJ/(mol.K)[/tex]

So, [tex]\Delta G^{0}=(-136.9kJ/mol)-(298K\times -0.1206\frac{kJ}{mol.K})[/tex]

              = -101.0 kJ/mol

So, [tex]\Delta G^{0}[/tex] at 298 K is -101.0 kJ/mol

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