Question

Use the ΔH°f and ΔH°rxn information provided to calculate ΔH°f for IF: ΔH°f (kJ/mol) IF7(g) + I2(g) → IF5(g) + 2 IF(g)

ΔH°rxn = -89 kJ

IF7(g) -941

IF5(g) -840

Answers:

(a) -190 KJ/mol

(b) 101 KJ/mol

(c) 24 KJ/mol

(d) -95 KJ/mol

(e) -146 KJ/mol

Answer 1

IF₇(g) + I₂(g) → IF₅(g) + 2 IF(g) , ΔH°_rxn = - 89 kJ

ΔH°rxn = (sum of ΔH°f products) - (sum of ΔH°f reactants)

given ΔH°rxn = -89 kJ

ΔH°f (IF7) = -941 kJ

ΔH°f(I₂) = 0 (elements in their standard state have zero ΔH°f)

ΔH°f(IF5) = -840 kJ
ΔH°f(IF) = ?

so ΔH°rxn = [ ΔH°f(IF5) + 2 X ΔH°f(IF) ] - [ ΔH°f(IF7) + ΔH°f(I₂) ]

-89 = [ -840 + 2 X ΔH°f(IF) ] - [ -941 + 0 ]

-89 = -840 + 2 X ΔH°f(IF) + 941

-89 = 101 + 2 X ΔH°f(IF)

2 X ΔH°f(IF) = -89 - 101 = -190

ΔH°f(IF) = -190/2 = -95

Answer is (d) -95 kJ/ mol