Calculate the enthalpy change for the following reaction :
C₂H₆O (g) + 3 O₂ (g) $\rightarrow$ 2 CO₂ (g) + 3 H₂O (g) , using the bond enthalpies given (in kJ mol^-1) :
C-H = + 412
C=O = + 742
C-O = + 358
C-C = + 347
O=O = + 498
O-H = + 464

Question

Calculate the enthalpy change for the following reaction :
C₂H₆O (g) + 3 O₂ (g)

\rightarrow

2 CO₂ (g) + 3 H₂O (g) , using the bond enthalpies given (in kJ mol^-1) :
C-H = + 412
C=O = + 742
C-O = + 358
C-C = + 347
O=O = + 498
O-H = + 464

Quizplus · Accepted Answer

We can calculate the overall enthalpy change by summing up the bonds broken minus the bonds formed. 
Bonds broken (Reactants): 
- 10 C-H bonds = - 4120 kJ/mol 
- 1 C=O bond = - 742 kJ/mol 
- 4 C-O bonds = - 1432 kJ/mol 
- 3 O=O bonds = - 1494 kJ/mol 
- 1 O-H bond = - 464 kJ/mol

Bonds formed (Products): 
- 4 C=O bonds = + 2968 kJ/mol 
- 6 C-H bonds = + 2472 kJ/mol 
- 3 O-H bonds = + 1392 kJ/mol

∆H = (Bonds broken) - (Bonds formed) 
∆H = (- 4120) - (742 + 1432 + 1494 + 464) - (- 2968 - 2472 - 1392) 
∆H = - 1029 kJ/mol

Therefore, the correct choice is D, which is - 1029 kJ/mol.

Calculate the Enthalpy Change for the Following Reaction : C2H6O →\rightarrow→

Calculate the Enthalpy Change for the Following Reaction :
C₂H₆O $\rightarrow$