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