Respuesta :
Answer:
The correct answer is option A.
Explanation:
[tex]PCl_3(g) + Cl_2(g)\rightarrow PCl_5(g)[/tex]
Enthalpy of formation of [tex]PCl_3[/tex] gas = [tex]\Delta H_f_{(PCl_3)}=-288.7 kJ/mol[/tex]
Enthalpy of formation of chlorine gas = [tex]\Delta H_f_{(Cl_2)}=-0 kJ/mol[/tex]
Enthalpy of formation of [tex]PCl_5[/tex] gas =[tex]\Delta H_f_{(PCl_5)}=-374.9 kJ/mol[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
For the given chemical reaction:
[tex]\Delta H_{rxn}=[(1\times \Delta H_f_{(PCl_5)})]-[(1\times \Delta H_f_{(PCl_3)})+(1\times \Delta H_f_{(Cl_2)})][/tex]
[tex]=[1\times (-374.9 kJ/mol)]-[1\times (-288.7 kJ/mol)+1\times 0 kJ/mol][/tex]
[tex]\Delta H_{rxn}=-86.2 kJ/mol[/tex]
Entropy of of [tex]PCl_3[/tex] gas = [tex]\Delta S_{(PCl_3)}=311.6 J/mol K[/tex]
Entropy of chlorine gas = [tex]\Delta S_{(Cl_2)}=223.1 J/mol K[/tex]
Entropy of [tex]PCl_5[/tex] gas =[tex]\Delta S_{(PCl_5)}=364.2 J/mol K[/tex]
The equation used to calculate enthalpy change is of a reaction is:
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta S_{rxn}=\sum [n\times \Delta S_(product)]-\sum [n\times \Delta S_(reactant)][/tex]
The equation for the entropy change of the above reaction is:
[tex]\Delta S_{rxn}=[(1\times \Delta S_{(PCl_5)})]-[(1\times \Delta S_{(PCl_3)})+(1\times \Delta S_{(Cl_2)})][/tex]
[tex]=[1\times 364.2 J/molK]-[1\times 311.6 J/mol K+1\times 223.1 J/mol K][/tex]
[tex]=-170.5 J/mol K[/tex]
[tex]\Delta G^o=\Delta H^o-T\Delta S^o[/tex]
[tex]\Delta G_{rxn}=\Delta H_{rxn}-T\Delta S_{rxn}[/tex]
At equilibrium , [tex]\Delta G_{rxn}=0[/tex]
[tex]\Delta G_{rxn}=\Delta H_{rxn}-T\Delta S_{rxn}[/tex]
[tex]\Delta H_{rxn}=T\Delta S_{rxn}[/tex]
[tex]T=\frac{\Delta H_{rxn}}{\Delta S_{rxn}}=\frac{-86.2 kJ/mol}{-170.5 J/mol}[/tex]
1 kJ = 1000 J
[tex]T=\frac{(-86.2\times 1000 J/mol)}{(-170.5 J/mol)}=505.57 k\approx 506 K[/tex]
At 506 K reaction will be at an equilibrium.
