Which balanced thermochemical equation corresponds to the enthalpy of formation (ΔHfo) of carbon monoxide, CO(g)? The enthalpy of formation (ΔHfo) of CO(g) is –110.53 kJ∙mol–1

Question

Which balanced thermochemical equation corresponds to the enthalpy of formation (ΔHfo) of carbon monoxide, CO(g)?  The enthalpy of formation (ΔHfo) of CO(g) is –110.53 kJ∙mol–1

BlackTom AI · Accepted Answer

Enthalpy of formation is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states. For CO(g), the elements are carbon in its standard solid form C(s) (graphite) and oxygen as O2(g).

1) Relevant knowledge points and concepts
- Standard enthalpy of formation ΔHf° refers to formation of exactly 1 mole of product from elements in their standard states.
- Standard states: C(s) for carbon, O2(g) for oxygen; ΔHf° of any pure element in its standard state is zero.
- Stoichiometry must produce one mole of the compound for the listed numerical ΔHf° to apply directly.

2) Step-by-step reasoning
- To form 1 mol CO(g) from elements: start with 1 mol C(s) + 1/2 mol O2(g) → 1 mol CO(g). This is stoichiometrically balanced and uses elements in standard states.
- The given ΔHf° = –110.53 kJ·mol–1 therefore corresponds exactly to this equation.

3) Intermediate calculations and stoichiometric checks
- Option 1: C(s) + ½O2(g) → CO(g) corresponds to formation of 1 mol CO; assigned ΔHf° = –110.53 kJ·mol–1 — consistent.
- Option 2: 2C(s) + O2(g) → 2CO(g) produces 2 mol CO. The enthalpy for that reaction would be 2 × (–110.53 kJ) = –221.06 kJ. Thus it cannot be paired with ΔHf° = –110.53 kJ·mol–1 without scaling; the given value is per mole, so option 2 is incorrect as written.
- Option 3: CO(g) + ½O2(g) → CO2(g) is oxidation of CO to CO2, not formation of CO from elements. Its enthalpy equals ΔHf°(CO2) − ΔHf°(CO), not simply ΔHf°(CO).
- Option 4: 2CO(g) + O2(g) → 2CO2(g) is just the doubled version of option 3; likewise not a formation reaction for CO.

4) Why the correct answer is correct
Option 1 is the only equation that represents the formation of exactly one mole of CO from its elemental forms in their standard states, matching the definition and the numerical value ΔHf° = –110.53 kJ·mol–1.

5) Why other options are incorrect (brief)
- Option 2: produces 2 mol CO so ΔH for that equation would be −221.06 kJ, not −110.53 kJ per the equation as written.
- Options 3 and 4: are oxidation reactions (CO → CO2) and not formation-from-elements reactions; they do not represent ΔHf° of CO.

6) Conclusion
The balanced thermochemical equation corresponding to ΔHf°(CO(g)) = –110.53 kJ·mol–1 is:
C(s) + ½O2(g) → CO(g) ΔHf° = –110.53 kJ·mol–1,
which matches the provided answer.

Which balanced thermochemical equation corresponds to the enthalpy of formation (ΔHfo) of carbon monoxide, CO(g)? The enthalpy of formation (ΔHfo) of CO(g) is –110.53 kJ∙mol–1单项选择题

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