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.

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answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error