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Given that ∆Hf° = 142.7 kJ/mol for O3(g) and ∆Hf° = 249.2 kJ/mol for O(g), calculate ∆H° for the reaction O2 (g) + O(g) O3(g).

User Asunez
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Final answer:

The enthalpy change for the reaction O2 (g) + O(g) → O3(g) is -106.5 kJ/mol, calculated using the standard enthalpy of formation values for O3(g) and O(g), and recognizing that ∆Hf° for O2 is zero.

Step-by-step explanation:

To calculate the enthalpy change (∆H°) for the reaction O2 (g) + O(g) → O3(g), we use the standard enthalpy of formation values provided for O3(g) and O(g).

The standard enthalpy of formation (∆Hf°) is the heat change when one mole of a substance is formed from its elements in their standard states.

In this case, we are given ∆Hf° = 142.7 kJ/mol for O3(g) and ∆Hf° = 249.2 kJ/mol for O(g).

The enthalpy change of the reaction (∆H°) is calculated using the formula:

∆H° = ∆Hf° (products) - ∆Hf° (reactants)

For the reaction under consideration:

∆H° = ∆Hf° (O3) - [∆Hf° (O2) + ∆Hf° (O)]

Where ∆Hf° (O2) is zero for elemental oxygen in its standard state. Thus:

∆H° = 142.7 kJ/mol (for O3) - [0 kJ/mol (for O2) + 249.2 kJ/mol (for O)]

∆H° = 142.7 kJ/mol - 249.2 kJ/mol

= -106.5 kJ/mol

Therefore, the enthalpy change for the formation of ozone from oxygen and atomic oxygen is -106.5 kJ/mol.

User Fyngyrz
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