Question

Estimate the enthalpy change for the combustion of one mole of acetylene, C2H2, to form carbon dioxide and water vapor. BE(C?H) = 456 kJ/mol) BE(C?C) = 962 kJ/mol BE(O=O) = 499 kJ/mol BE(C=O) = 802 kJ/mol BE(O?H) = 462 kJ/molAnswer choices:A. +653 kJ/molB. ?155 kJ/molC. +1010 kJ/molD. ?1759 kJ/molE. ?1010 kJ/mol

Answer 1

Answer:

ΔH rxn =  -1010 kJ/molC₂H₂

Explanation:

To obtain the enthalpy change for a reaction from bond energies what we do is to make an inventory of the bonds broken and formed for the balanced chemical reaction:

C₂H₂ + 5/2O₂   ⇒   2CO₂ + H₂O

Bond Broken                                    Bonds Formed

2 C-H + 1 C≡C + 5/2 O=O                4C=O + 2 H-O

Enthalpy bonds broken:

2 mol (456 kJ/mol)+ 1 mol (962 kJ/mol) + 5/2 mol (499 kJ/mol)  =  3121.5 kJ

Enthalpy bond formed:

4 mol (802 kJ/mol) + 2 mol (462 kJ/mol) = 4132.0 kJ

ΔH rxn = H broken - H formed =  3121.5 kJ - 4132.0 kJ = - 1010  kJ (per mol C₂H₂ )

Answer 2

The enthalpy change will be "-1010 kJ/mol".

According to the question,

The combustion of acetylene occur as:

• $H-C \equiv C-H+\frac{5}{2} O=O \rightarrow 2 O =C =O+H-O-H$

here,

Breaking bonds:

• $2C-H$
• $1C \equiv C$

• $\frac{5}{2} O=O$

Formed bonds:

• $4C=O$
• $2O-H$

As we know,

→ $Enthalpy \ change = \Sigma H_{Broken} + \Sigma H_{Formed}$

By substituting the values, we get

                               $= [2(C-H)+1(C \equiv C)+\frac{5}{2}(O =O) ]+ [4(C=O)+2 (O-H)]$

$= 2(456)+962+\frac{5}{2}(499)+4(-802)+2(-462)$

$=912+962+1247.5-3208-924$

$= -1010 \ kJ/mol$

Thus the above answer is right.

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