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

Question

contestada

Respuesta :

nwandukelechi nwandukelechi · Answer 1 · 2020-02-13T09:00:44+01:00

Answer:

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.