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ΔH for H₂ + 1/2O₂ -> H₂O

A. -285.8 kJ/mol
B. -57.9 kJ/mol
C. 285.8 kJ/mol
D. 57.9 kJ/mol

User Arlind
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1 Answer

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Final Answer:

The ΔH for the reaction H₂ + 1/2O₂ -> H₂O is -285.8 kJ/mol (Option c).

Step-by-step explanation:

The given value of -285.8 kJ/mol represents the enthalpy change (ΔH) for the reaction H₂ + 1/2O₂ -> H₂O. The negative sign indicates an exothermic reaction, where energy is released to the surroundings. This value is often expressed in kJ/mol to represent the heat evolved or absorbed per mole of reactant (Option c).

The enthalpy change in a chemical reaction is a measure of the heat energy exchanged with the surroundings. In this case, -285.8 kJ/mol signifies that 285.8 kilojoules of energy are released for every mole of H₂ reacting with 1/2 mole of O₂ to produce H₂O. The reaction involves the formation of water, and the negative value indicates a release of energy, making it exothermic.

Understanding the sign and magnitude of ΔH is crucial in assessing the thermodynamic nature of a reaction. In this context, the negative value signifies an exothermic process, and the specific quantity, -285.8 kJ/mol, provides a quantitative measure of the heat released during the formation of water from hydrogen and oxygen.

User Alexandru Dinu
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