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What is the standard enthalpy change for the reaction 2H₂(g) + O₂(g) -> 2H₂O(l), given the standard enthalpies of formation for H₂O(l), H₂(g) and O₂(g) as -285.8 kJ/mol, 0 kJ/mol and 0 kJ/mol, respectively?
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Mar 1
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Chemical thermodynamics
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TheScientist
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What is the standard enthalpy change for the reaction 2H₂(g) + O₂(g) -> 2H₂O(l), given the standard enthalpies of formation for H₂O(l), H₂(g) and O₂(g) as -285.8 kJ/mol, 0 kJ/mol and 0 kJ/mol, respectively?
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To calculate the standard enthalpy change for the reaction, we can use the formula:H = Hf products - Hf reactants where H is the standard enthalpy change for the reaction, and Hf is the standard enthalpy of formation for each substance.For the given reaction:2H g + O g -> 2HO l The standard enthalpies of formation are:Hf HO l = -285.8 kJ/molHf H g = 0 kJ/molHf O g = 0 kJ/molNow, we can plug these values into the formula:H = [2 -285.8 kJ/mol ] - [2 0 kJ/mol + 1 0 kJ/mol ]H = -571.6 kJ/mol - 0 kJ/mol H = -571.6 kJ/molThe standard enthalpy change for the reaction 2H g + O g -> 2HO l is -571.6 kJ/mol.
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