Question

Xenon forms 3 different Fluorides with $F_2$ gas.
Look at the following reaction:
$Xe\left(g\right)+2F_2\left(g\right)\to Xe{F}_4\left(s\right)\Delta G$<$0$
and choose theoretically appropriate arguments assuming the reaction is carried out in a sealed tube at a constant temperature T:

• A. The change in entropy for the given reaction is negative
• B. The above reaction is endothermic
• C. The reaction has a negative enthalpy change
• D. The (magnitude of ) is greater than the (magnitude of )

Answer 1

Answer: (A), (C), (D)

The correct options are
A The change in entropy for the given reaction is negative
C The reaction has a negative enthalpy change
D The (magnitude of ) is greater than the (magnitude of )

Let us break down this problem one step at a time. It is given that
$Xe\left(g\right)+2F_2\left(g\right)\to Xe{F}_4\left(s\right)\Delta G$<$0$

The reactants are all gaseous while the product is solid. Definitely $\Delta S$<$0$
This implies that $T\Delta S$<$0.Or-T\Delta S$>$0$
For any reaction to be thermodynamically feasible, $\Delta G$<$0$
But $\Delta G=\Delta H-T\Delta S.So\Delta H-T\Delta S$<$0$
Now we have established that - $T\Delta S$ is a positive quantity (since $\Delta S$ is negative). So the above equation is of the form, $\Delta H+\left(apositivequantity\right)$<$0$.
Clearly for $\Delta G$ to be negative, $\Delta H$<$0$ such that:
magnitude of $\Delta H$ > magnitude of the quantity $T.\Delta S$