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6 July, 15:19

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

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  1. 6 July, 15:31
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    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.
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