Question

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

Answer 1

Answer:

$\frac{1C}{2} +F -----> A+B+D$       ;ΔH = + 45.6 kJ/mol

Explanation:

From the question; suppose the first equation is reversed and multiplied by $\frac{1}{6}$; we have:

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

$\frac{3D}{6} ----->\frac{3A}{6}+\frac{6B}{6}$                  ; Δ$H = +\frac{402}{6} kJ/mol$

$\frac{1D}{2} ----->\frac{1A}{2}+B$                    ; Δ$H = +67.0kJ/mol$ ----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

$\frac{1E}{2} +\frac{2F}{2}------>\frac{1A}{2}$            ;ΔH = $\frac{-108.3 kJ/mol}{2}$

$\frac{1E}{2}+F----->\frac{1A}{2}$                  ;ΔH = -54.15 kJ/mol --- Equation 2

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

$\frac{C}{2}----->\frac{E}{2}+\frac{3D}{2}$                   ;ΔH =   $\frac{+65.5kJ/mol}{2}$

$\frac{1C}{2} ----->\frac{1E}{2}+\frac{3D}{2}$                ;ΔH = + 32.75 kJ/mol --- Equation 3

Adding these three  adjusted equations; we have:

$\frac{1D}{2} ----->\frac{1A}{2}+B$                    ; Δ$H = +67.0kJ/mol$

$\frac{1E}{2}+F----->\frac{1A}{2}$                    ;ΔH = -54.15 kJ/mol

$\frac{1C}{2} ----->\frac{1E}{2}+\frac{3D}{2}$                  ;ΔH = + 32.75 kJ/mol

                                                                                                       

$\frac{1C}{2} +F -----> A+B+D$       ;ΔH = + 45.6 kJ/mol

                                                                                                       

∴ The net reaction = $\frac{1C}{2} +F -----> A+B+D$       ;ΔH = + 45.6 kJ/mol