Answer:
The value of ΔG° at 201.0∘C for the formation of phosphorous trichloride is -595.5 kJ/mol.
Explanation:
[tex]P_2(g)+3Cl_2(g)\rightarrow 2PCl_3(g)[/tex]
Gibbs free energy of the reaction = ΔG° = -642.9 kJ/mol = -642,900 J/mol
Enthalpy of the reaction = ΔH° = -720.5 kJ/mol = -720,500 J/mol
Change in entropy of the reaction = ΔS° = -263.7 J/mol
Gibbs free energy of the reaction at 201.0°C = ΔG° = ?
T = 201.0°C = 474.15 K
Using Gibbs's energy equation:
[tex]\Delta G^o=\Delta H^o-T\Delta S^o[/tex]
[tex]=-720,500 J/mol- 474.15 K\times (-263.7 J/mol)[/tex]
[tex]\Delta G^o=-595,466.64 J/mol=-595.5 kJ/mol[/tex]
The value of ΔG° at 201.0∘C for the formation of phosphorous trichloride is -595.5 kJ/mol.
The value of ΔG° at 201.0°C is mathematically given as
ΔG° = −595.5kJ/mol
Question Parameter(s):
The value of ΔG∘ at 201.0∘C for the formation of phosphorous trichloride
At 25.0∘C for this reaction,
ΔH∘ is −720.5kJ/mol,
ΔG∘ is −642.9kJ/mol
ΔS∘ is −263.7J/K.
Generally, the equation for Gibbs's energy is mathematically given as
[tex]\Delta G^o=\Delta H^o-T\Delta S^o[/tex]
ΔG° = −720500 −474.15x(−263.7)
ΔG° = −595.5kJ/mol
In conclusion, the value of ΔG° at 201.0°C is
ΔG° = −595.5kJ/mol
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