QQuestionChemistry
QuestionChemistry
The standard molar enthalpy of formation for oxygen gas, O
2
(g), is zero. The enthalpy of formation for atomic oxygen, O(g), is + 250 kJ/mol. Why is there a difference?
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Answer
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Step 1:I'll solve this problem step by step, explaining the reasoning behind the difference in standard molar enthalpies.
Step 2:: Understand the Given Information
- Standard molar enthalpy of formation for O₂(g) = 0 kJ/mol - Standard molar enthalpy of formation for O(g) = + 250 kJ/mol
Step 3:: Define Standard Enthalpy of Formation
The standard enthalpy of formation ($$\Delta H^{\circ}_{f}$$) is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states.
Step 4:: Analyze O₂(g) Formation
Since the molecule is already in its standard state, $$\Delta H^{\circ}_{f} = 0$$ kJ/mol
For molecular oxygen (O₂), the formation reaction is:
Step 5:: Analyze O(g) Formation
\mathrm{O_{2}}(g) \rightarrow 2\mathrm{O}(g)
For atomic oxygen, the formation reaction involves breaking the O₂ molecule:
Step 6:: Explain the Energy Difference
The positive enthalpy (+ 250 kJ/mol) represents the energy required to break the oxygen-oxygen double bond in the O₂ molecule. This bond dissociation energy explains why atomic oxygen has a higher enthalpy of formation.
Step 7:: Understand Bond Breaking
- O₂ molecules have a strong double bond - Breaking this bond requires significant energy input - This energy input is reflected in the positive enthalpy of atomic oxygen
Final Answer
The difference in standard molar enthalpies (0 kJ/mol for O₂ vs. + 250 kJ/mol for O) is due to the energy required to break the strong oxygen-oxygen double bond, converting molecular oxygen into individual oxygen atoms.
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