Question

For the gaseous reaction of xenon (Xe) and fluorine (F) to form xenon hexafluoride (XeF₆), ΔH° = -402 kJ/mol and ΔG° = -280 kJ/mol at 298 K.

Xe (g) + 3F₂ (g) → XeF₆ (g)

a. Calculate the ΔS° for the reaction.

b. Calculate the ΔG° at 500 K.

c. Calculate the K_eq values for the reaction at 298 K and 500 K.

d. At what temperature value does the reaction become non-spontaneous?

Answer 1

Xe (g) + 3F₂ (g) → XeF₆ (g)

a) Given,

ΔH° = -402 kJ/mol

ΔG° = -280 kJ/mol

T= 298 K.

Formula

ΔS° = 409 J

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b) Assuming ΔH° and ΔS° wont change with temperature

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c) formula to calculate Keq

At 298 K

ΔG° = -280 kJ/mol = -280 x 1000 J/mol

at 298 K

At 500 K

ΔG° = -198 kJ/mol = -198 x 1000 J/mol

at 500 K

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d) If ΔG° = 0, then it is the limit of reaction spontaneity

so when ΔG° = 0,

adding 402000 on both sides

T = 983 K

At 983 K the reaction will become non-spontaneous