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?

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

a graduated cylinder.

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