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

Respuesta :

Answer:

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

Explanation:

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

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

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

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

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

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

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

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

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

Adding these three  adjusted equations; we have:

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

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

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

                                                                                                       

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

                                                                                                       

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

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