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?3 A + 6 B → 3 D, ΔH = -402 kJ/molE + 2 F → A, ΔH = -108.3 kJ/molC → E + 3 D, ΔH = +65.5 kJ/mol

      ;ΔH = + 45.6 kJ/mol

Explanation:

From the question; suppose the first equation is reversed and multiplied by ; we have:

3D       ------>       3A +      6B          ; ΔH   =  + 402 kJ/mol

                 ; Δ

                   ; Δ ----Equation 1

The second and third equations are divided by 2; so are going to have the following:

E     +     2 F -------> A                     ;ΔH = -108.3 kJ/mol

           ;ΔH =

                 ;ΔH = -54.15 kJ/mol --- Equation 2

C -------> E     +    3 D                     ;ΔH = +65.5 kJ/mol

                  ;ΔH =  

               ;ΔH = + 32.75 kJ/mol --- Equation 3

Adding these three  adjusted equations; we have:

                   ; Δ

                   ;ΔH = -54.15 kJ/mol

                 ;ΔH = + 32.75 kJ/mol

      ;ΔH = + 45.6 kJ/mol

∴ The net reaction =       ;ΔH = + 45.6 kJ/mol