Question

Find the change in enthalpy (ΔH) for the reactions 2 H2 (g) + O2(g) --> 2 H2O (g)

Answer 1

The change in enthalpy is -808 kJ.

Step-by-step explanation:

The given information from the exercise is:

C-H bonds energy: 413 kJ

O=O bonds energy: 495 kJ

C=O bonds energy: 799 kJ

H-O bonds energy: 463 kJ

Chemical reaction:

`CH_4+2O_2\rightarrow CO_2+2H_2O`

1st) It is necessary to calculate the energy of bonds broken:

4 x (C-H) = 4 x 413 kJ = 1652 kJ

2 x (O=O) = 2 x 495 kJ = 990 kJ

Total energy of bonds broken: 1652 kJ + 990 kJ = 2642 kJ

2nd) Now we have to calculate the energy of bonds made:

2 x (C=O) = 2 x 799 kJ = 1598 kJ

4 x (H-O) = 4 x 463 kJ = 1852 kJ

Total energy of bonds made: 1598 kJ + 1852 kJ = 3450 kJ

3rd) Finally, we can calculate the enthalpy for the reaction subtracting the Total energy of bonds broken minus the Total energy of bonds made:

Change in enthalpy = Total energy of bonds broken - Total energy of bonds made

Change in enthalpy = 2642 kJ - 3450 kJ

Change in enthalpy = -808 kJ

So, the change in enthalpy is -808 kJ.