Estimate δh for the reaction 6co2(g)+6h2o(l)→ c6h12o6(aq)+6o2(g) using bond dissociation energies

Bonds Broken

6CO2:

C-O = 358
358 x 12 = 4296 kJ/mol

6H20:

H-O = 459
459 x 12 = 5508 kJ/mol

Total Energy Absorbed:

4296 + 5508 = 9804 kJ/mol

Bonds Formed

C6H1206:

5 C-C bonds
C-C = 346
346 x 5 = 1730kJ/mol

7 H-C bonds
H-C = 411
411 x 7 = 2877 kJ/mol

C=O = 799 kJ/mol

3 C-O bonds
C-O = 358
358 x 3 = 1074 kJ/mol

5 O-H bonds
O-H = 459
459 x 5 = 2295 kJ/mol

1730+2877+799+1074+2295 = 8775 kJ/mol

602:

O=O = 494
494 x 6 = 2964 kJ/mol

Total Energy Released:

8775 + 2964 = 11739 kJ/mol

Total Enthalpy Change

9804 - 11739 = -1934 kJ/mol

ΔH = - 1934 kJ/mol

(This took me a while, but it was really fun. I learnt how to calculate bond energies for the first time yesterday, so this was good practice).

Mgo2 is the oxidation state of manganese oxide

29.1-12=17.1the answer is this one