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

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) .

A) -36.66 kJ/mol
B) -46.66 kJ/mol
C) -56.66 kJ/mol
D) -66.66 kJ/mol
E) None of the above

Correct Answer:

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