Question

At what temperature is the following reaction at equilibrium when all substances are at standard pressure?

Assume that entropies and enthalpies of reaction do not vary with temperature.
PCl3(g) + Cl2(g) PCl5(g) Substance ∆Hºf , kJ mol–1 Sº, J mol–1 K –1 PCl3(g) –288.7 311.6 Cl2(g) 0 223.1 PCl5(g) –374.9 364.2 (A) 506 K (B) 1640 K (C) 1980 K (D) 4260 K

Answer 1

The correct answer is option A.

Step-by-step explanation:

`PCl_3(g) + Cl_2(g)\rightarrow PCl_5(g)`

Enthalpy of formation of
`PCl_3` gas =
`\Delta H_f_((PCl_3))=-288.7 kJ/mol`

Enthalpy of formation of chlorine gas =
`\Delta H_f_((Cl_2))=-0 kJ/mol`

Enthalpy of formation of
`PCl_5` gas =
`\Delta H_f_((PCl_5))=-374.9 kJ/mol`

The equation used to calculate enthalpy change is of a reaction is:

`\Delta H_(rxn)=\sum [n* \Delta H_f(product)]-\sum [n* \Delta H_f(reactant)]`

For the given chemical reaction:

`\Delta H_(rxn)=[(1* \Delta H_f_((PCl_5)))]-[(1* \Delta H_f_((PCl_3)))+(1* \Delta H_f_((Cl_2)))]`

`=[1* (-374.9 kJ/mol)]-[1* (-288.7 kJ/mol)+1* 0 kJ/mol]`

`\Delta H_(rxn)=-86.2 kJ/mol`

Entropy of of
`PCl_3` gas =
`\Delta S_((PCl_3))=311.6 J/mol K`

Entropy of chlorine gas =
`\Delta S_((Cl_2))=223.1 J/mol K`

Entropy of
`PCl_5` gas =
`\Delta S_((PCl_5))=364.2 J/mol K`

The equation used to calculate enthalpy change is of a reaction is:

The equation used to calculate enthalpy change is of a reaction is:

`\Delta S_(rxn)=\sum [n* \Delta S_(product)]-\sum [n* \Delta S_(reactant)]`

The equation for the entropy change of the above reaction is:

`\Delta S_(rxn)=[(1* \Delta S_((PCl_5)))]-[(1* \Delta S_((PCl_3)))+(1* \Delta S_((Cl_2)))]`

`=[1* 364.2 J/molK]-[1* 311.6 J/mol K+1* 223.1 J/mol K]`

`=-170.5 J/mol K`

`\Delta G^o=\Delta H^o-T\Delta S^o`

`\Delta G_(rxn)=\Delta H_(rxn)-T\Delta S_(rxn)`

At equilibrium ,
`\Delta G_(rxn)=0`

`\Delta G_(rxn)=\Delta H_(rxn)-T\Delta S_(rxn)`

`\Delta H_(rxn)=T\Delta S_(rxn)`

`T=(\Delta H_(rxn))/(\Delta S_(rxn))=(-86.2 kJ/mol)/(-170.5 J/mol)`

1 kJ = 1000 J

`T=((-86.2* 1000 J/mol))/((-170.5 J/mol))=505.57 k\approx 506 K`

At 506 K reaction will be at an equilibrium.