Answer:
F + C/2 → A + B +D, ΔH +400 kJ/mol
Explanation:
3 A + 6 B → 3 D, ΔH = -446 kJ/mol
first equation is reversed and multiplied by 1/6
1/6(3 D → 3 A + 6 B) , ΔH = +(446) kJ/mol
( D/2 → A/2 + B) , ΔH = +(446) kJ/mol
E + 2 F → A, ΔH = -107.9 kJ/mol
second equation is divided by 2,
(E/2 + F → A/2), ΔH = -107.9 kJ/mol
C → E + 3 D, ΔH = +61.9 kJ/mol
third equation is divided by 2,
(C/2 → E/2 + 3/2 D), ΔH = +61.9 kJ/mol
the three adjusted equations are added.
( D/2 → A/2 + B) + (E/2 + F → A/2)+ (C/2 → E/2 + 3/2 D), ΔH = +61.9 kJ/mol + -107.9 kJ/mol +(446) kJ/mol
D/2 +E/2 + F+ C/2 → A/2 + B +A/2 +E/2 + 3/2 D , ΔH +400 kJ/mol
F + C/2 → A + B +D, ΔH +400 kJ/mol
