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

F + C/2 → A + B +D, ΔH +400 kJ/mol

Step-by-step 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