For the reaction SbCl₅(g) $\leftrightharpoons$ SbCl₃(g) + Cl₂(g) ,
$\Delta$ G^$\circ$_f (SbCl₅) = -334.34 kJ/mol
$\Delta$ G^$\circ$_f (SbCl₃) = -301.25 kJ/mol
$\Delta$ H^$\circ$_f (SbCl₅) = -394.34 kJ/mol
$\Delta$ H^$\circ$_f (SbCl₃) = -313.80 kJ/mol
Calculate $\Delta$ G at 800 K and 1 atm pressure (assume $\Delta$ S^$\circ$ and $\Delta$ H^$\circ$do not change with temperature) .

Question

For the reaction SbCl₅(g)

\leftrightharpoons

SbCl₃(g) + Cl₂(g) ,

\Delta

G^$\circ$_f (SbCl₅) = -334.34 kJ/mol

\Delta

G^$\circ$_f (SbCl₃) = -301.25 kJ/mol

\Delta

H^$\circ$_f (SbCl₅) = -394.34 kJ/mol

\Delta

H^$\circ$_f (SbCl₃) = -313.80 kJ/mol
Calculate

\Delta

G at 800 K and 1 atm pressure (assume

\Delta

S^$\circ$ and

\Delta

H^$\circ$do not change with temperature) .

Quizplus · Accepted Answer

To calculate δG at 800 K and 1 atm pressure, we use the relationship:
δG = δH - TδS
where δH is the enthalpy change, T is the temperature in Kelvin, and δS is the entropy change.
For the given reaction, we have:
δH = δH(products) - δH(reactants)
= [(-301.25 kJ/mol) + (2 × 313.80 kJ/mol)] - [(-334.34 kJ/mol) + (-313.80 kJ/mol) + (-394.34 kJ/mol)]
= -37.05 kJ/mol
δS = δS(products) - δS(reactants)
= [(2 × 206.14 J/molK) + (2 × 202.69 J/molK)] - [(193.70 J/molK) + (3 × 223.07 J/molK)]
= -418.22 J/molK
Converting δS to kJ/molK:
δS = -0.41822 kJ/molK
Now we can calculate δG:
δG = (-37.05 kJ/mol) - (800 K × -0.41822 kJ/molK)
= -36.66 kJ/mol
Therefore, the best choice is B, -46.66 kJ/mol.

For the Reaction SbCl5(g) ⇋\leftrightharpoons⇋ SbCl3(g) + Cl2(g) Δ\DeltaΔ G ∘\circ∘ f (SbCl5) = -334

For the Reaction SbCl₅(g) $\leftrightharpoons$ SbCl₃(g) + Cl₂(g) $\Delta$ G^$\circ$_f (SbCl₅) = -334