Question

For the following reaction:



IBr(g) + 4F₂(g) → IF₅(g) + BrF₃(g)

Compound		ΔH°f (kJ mol-1)		S° (J mol-1 K-1)
IBr (g)		40.88		258.95
F2 (g)		0.00		202.80
IF5 (g)		-840.31		334.50
BrF3 (g)		-255.59		292.30

Determine the temperature (to two decimal places in K) such that the reaction is in equilibrium in its standard states.

Answer 1

Step 1:

Given:

Hof(IBr(g)) = 40.88 KJ/mol

Hof(F2(g)) = 0.0 KJ/mol

Hof(IF5(g)) = -840.31 KJ/mol

Hof(BrF3(g)) = -255.59 KJ/mol

Balanced chemical equation is:

IBr(g) + 4 F2(g) ---> IF5(g) + BrF3(g)

ΔHo rxn = 1*Hof(IF5(g)) + 1*Hof(BrF3(g)) - 1*Hof( IBr(g)) - 4*Hof(F2(g))

ΔHo rxn = 1*(-840.31) + 1*(-255.59) - 1*(40.88) - 4*(0.0)

ΔHo rxn = -1136.78 KJ

Step 2:

Given:

Sof(IBr(g)) = 258.95 J/mol.K

Sof(F2(g)) = 202.8 J/mol.K

Sof(IF5(g)) = 334.5 J/mol.K

Sof(BrF3(g)) = 292.3 J/mol.K

Balanced chemical equation is:

IBr(g) + 4 F2(g) ---> IF5(g) + BrF3(g)

ΔSo rxn = 1*Sof(IF5(g)) + 1*Sof(BrF3(g)) - 1*Sof( IBr(g)) - 4*Sof(F2(g))

ΔSo rxn = 1*(334.5) + 1*(292.3) - 1*(258.95) - 4*(202.8)

ΔSo rxn = -443.35 J/K

Step 3:

At equilibrium, ΔGo = 0.0 KJ/Mol

ΔGo = 0.0 KJ/mol

ΔHo = -1136.78 KJ/mol

ΔSo = -443.35 J/mol.K

= -0.44335 KJ/mol.K

use:

ΔGo = ΔHo - T*ΔSo

0.0 = -1136.78 - T *-0.4434

T = 2564.07 K

Answer: 2564.07 K