Use the provided reactions and their changes in enthalpy H2O (ℓ) ←→ H2 (g) + 1 2 O2 (g) ∆H = 285.8 kJ · mol-1 CH4 (g) + 2 O2 (g) ←→ CO2 (g) + 2 H2O (ℓ) ∆H = - 890.0 kJ · mol-1 CH4 (g) ←→ Cgraphite (s) + 2 H2 (g) ∆H = 74.9 kJ · mol-1 to find the change in enthalpy for the reaction CO2 (g) ←→ Cgraphite (s) + O2 (g) 1. 679.1 kJ · mol-1 2. - 227.2 kJ · mol-1 3. - 529.3 kJ · mol-1

4. 393.3 kJ/mol.

Explanation:

1. H2O (ℓ) → H2 (g) + 1/2O2

∆H = 285.8 kJ/mol.

2. CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (ℓ)

∆H = - 890.0 kJ/mol.

3. CH4 (g) → Cgraphite (s) + 2H2 (g)

∆H = 74.9 kJ/mol.

Flipping and then multiplying Equation 1. by 2,

2H2 (g) + O2 → 2 H2O (ℓ)

∆H = - 571.6 kJ/mol.

Note: ∆H sign changes.

CO2 (g) + 2 H2O (ℓ) → CH4 (g) + 2O2 (g)

∆H = 890.0 kJ/mol.

Adding all the equations and enthalpies together,

2H2 (g) + O2 → 2H2O (ℓ)

CO2 (g) + 2H2O (ℓ) → CH4 (g) + O2 (g)

CH4 (g) → Cgraphite (s) + 2H2 (g)

= CO2 (g) → Cgraphite (s) + O2 (g)

= (-571.6 + 890.0 + 74.9)

= 393.3 kJ/mol.