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Consider the following equations: 3A + 6B â 3D ÎH = â418 kJ E + 2F â A ÎH = â118.2 kJ C â E + 3D ÎH = â60.8 kJ 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 and what is the overall heat of this reaction?

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Answer:

The net reaction is:

1/2 C    +    F  +               →      A   +   B       D  

The overall heat of this reaction = 40.96 kJ

Explanation:

Given that:

3A   +   6B   →   3D      ΔH = -418  kJ

E       +  2F   →    A       ΔH =  -118.2 kJ

             C    →  E  +   3D ΔH = 60.8  kJ

The first process says we should reverse the first equation and multiply it by 1/6

i.e.

3D × 1/6 →  3A × 1/6 + 6B ×  1/6   ;  ΔH  =  +1/6 × 418 kJ

1/2D →   1/2 A  + B                         ; ΔH  =  69.66 kJ

Then the second and the third equation are divided by 2

1/2 × E   + F → 1/2 A           ; ΔH  = -1/2  ×  118.2  kJ

1/2C  →   1/2 E   + 3/2  D    ; ΔH  =  1/2  ×  60.8 kJ

By adjusting the three equations together; we have:

1/2D →   1/2 A  + B             ; ΔH  =  69.66 kJ

1/2 × E   + F → 1/2 A           ; ΔH  =   -59.1  kJ

1/2C  →   1/2 E   + 3/2  D    ; ΔH  =   30.4  kJ                      

1/2C + F → A + B + D         ; ΔH  =  (69.66 - 59.1 + 30.4) kJ

                                            ΔH  =  40.96  kJ

The net reaction is:

1/2 C    +    F  +               →      A   +   B       D  

The overall heat of this reaction = 40.96 kJ

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