Question

Using the bond energies provided below, calculate ?H° for the reaction

CH4(g) + 4Cl2(g) ? CCl4(g) + 4HCl(g)
Bond energies:
C–H = 413 kJ/mol
Cl–Cl = 243 kJ/mol
C–Cl = 339 kJ/mol,
H–Cl = 427 kJ/mol

Answer 1

ΔHºrxn = - 440 kJ

Step-by-step explanation:

What we need to remember in this question is that given the bond energies for a reaction, the enthalpy change , ΔHº , is given by the expression:

ΔHºrxn = ∑ energies of the bonds broken - ∑ energies of the bonds formed.

where s ∑ is summatory.

The reaction is CH₄(g) + 4 Cl₂(g) ⇒ CCl₄(g) + 4 HCl(g)

So lets make the inventory of the bond broken and bonds formed

Bonds Broken:

4 C-H = 4 mol (413 kJ/mol ) = 1652 kJ

4 Cl-Cl = 4 mol ( 243 kJ/mol = 972 kJ/mol

Bond Formed:

4 C-Cl = 4 mol (339 kJ/mol ) = 1356 kJ

4 H-Cl = 4 mol (427 kJ/mol ) = 1708 kJ

Now we can calculate ΔHºrxn :

ΔHºrxn = ∑ energies of the bonds broken - ∑ energies of the bonds formed

ΔHºrxn = ( 1652 + 972 ) kJ - ( 1356 + 1708 ) kJ

= 2624 kj - 3064 kJ

= - 440 kJ