For the reaction $$C_2H_6 \to C_2H_4 + H_2$$ the reaction enthalpy $$\Delta_r H$$ in kJ mol$$^{-1}$$ is ________. (Round off to the Nearest Integer).
[Given: Bond enthalpies in kJ mol$$^{-1}$$: C-C: 347, C=C: 611; C-H: 414, H-H: 436]

We use the bond enthalpy method: $$\Delta_r H = \sum(\text{Bond enthalpies of bonds broken}) - \sum(\text{Bond enthalpies of bonds formed})$$.

In the reaction $$\text{C}_2\text{H}_6 \rightarrow \text{C}_2\text{H}_4 + \text{H}_2$$, we break all bonds in ethane and form all bonds in ethylene and hydrogen.

Bonds broken in $$\text{C}_2\text{H}_6$$: one C-C bond and six C-H bonds. Energy required $$= 1 \times 347 + 6 \times 414 = 347 + 2484 = 2831$$ kJ/mol.

Bonds formed in $$\text{C}_2\text{H}_4$$ and $$\text{H}_2$$: one C=C bond, four C-H bonds, and one H-H bond. Energy released $$= 1 \times 611 + 4 \times 414 + 1 \times 436 = 611 + 1656 + 436 = 2703$$ kJ/mol.

Therefore $$\Delta_r H = 2831 - 2703 = 128$$ kJ/mol.