Respuesta :
ΔG⁰ = ΔH⁰ - T ΔS⁰
ΔG⁰ : Standard free energy of formation of acetylene
ΔH⁰ : Standard enthalpy of formation (226.7 kJ/mol)
ΔS⁰ : Standard entropy change (58.8 J / K. mol)
T : Temperature 25°C = 298 K (room temperature)
ΔG⁰ = 226.7 - (298 x 58.8 x 10⁻³) = 209.2 kJ /mol
ΔG⁰ : Standard free energy of formation of acetylene
ΔH⁰ : Standard enthalpy of formation (226.7 kJ/mol)
ΔS⁰ : Standard entropy change (58.8 J / K. mol)
T : Temperature 25°C = 298 K (room temperature)
ΔG⁰ = 226.7 - (298 x 58.8 x 10⁻³) = 209.2 kJ /mol
Explanation:
The given data is as follows.
[tex]\Delta H^{o}[/tex] = 226.7 kJ/K mol,
[tex]\Delta S^{o}[/tex] = 58.8 J/mol = [tex]58.8 \times 10^{-3} kJ/K mol[/tex]
T = [tex]25^{o}C[/tex] = (25 + 273) K = 298 K
Now, calculate the standard free energy of formation of acetylene ([tex]C_{2}H_{2}[/tex]) as follows.
[tex]\Delta G^{o} = \Delta H^{o} - T \Delta S^{o}[/tex]
= 226.7 - kJ/mol - 298 K \times 58.8 \times 10^{-3} kJ/K mol[/tex]
= 209.2 kJ/mol
Thus, we can conclude that the standard free energy of formation of acetylene is 209.2 kJ/mol.