1. Bond energy calculation

2. Use the bond energy data on the right to estimate the ∆H for this reaction: CH 4 (g) + 2Cl 2 (g) → CH 2 Cl 2 (g) + 2HCl(g) Cl–Cl H–Cl H H H HC –– – – H Cl HC –– – – H–Cl 431 kJ mol -1 Cl–Cl 243 kJ mol -1 C–Cl 327 kJ mol -1 H–C 414 kJ mol -1 1 Rewrite the equation, showing every bond. 2 List each bond broken and made, together with its bond energy. Bond breaking (∆H positive): 4 × H–C = 4 × 414 kJ mol -1 2 × Cl–Cl = 2 × 243 kJ mol -1 Bond making (∆H negative): 2 × H–C = 2 × –414 kJ mol -1 2 × C–Cl = 2 × –327 kJ mol -1 2 × H–Cl = 2 × –431 kJ mol -1

3. Bond breaking: 4 × H–C = 4 × 414 kJ mol -1 2 × Cl–Cl = 2 × 243 kJ mol -1 Bond making: 2 × H–C = 2 × –414 kJ mol -1 2 × C–Cl = 2 × –327 kJ mol -1 2 × H–Cl = 2 × –431 kJ mol -1 3 Find the sum of each column. 2142 kJ mol -1 –2344 kJ mol -1 4 Combine the energy absorbed in bond breaking and the energy released in bond making to find the ∆H. ∆H = 2142 kJ mol -1 – 2344 kJ mol -1 = -202 kJ mol -1 We could simplify by removing two H–C bonds from each side.

4. Common trap Pay attention to single and multiple bonds: A single C=O bond is not the same as two C-O bonds. Similarly, there is a difference between N ≡ N and N-N.