Question

Consider the following equations. 3 A + 6 B → 3 D, ΔH = -446 kJ/mol E + 2 F → A, ΔH = -107.9 kJ/mol C → E + 3 D, ΔH = +61.9 kJ/mol Suppose the first equation is reversed and multiplied by 1/6, the second and third equations are divided by 2, and the three adjusted equations are added. What is the net reaction? chemPadHelp

Answer 1

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