asked 49.8k views
4 votes
If the reaction, 2 H2 + O2 --> 2 H2O, releases 483.6 kJ, then the standard enthalpy of formation of H2O = _______ kJ/mole.

A. -483.6
B. +483.6
C. -241.8
D. +241.8

1 Answer

2 votes

Final answer:

The standard enthalpy of formation of H2O can be determined using the enthalpy change of the reaction and stoichiometry. Given that the reaction 2 H2 + O2 -> 2 H2O releases 483.6 kJ, the standard enthalpy of formation of H2O is determined to be 241.8 kJ/mol.

Step-by-step explanation:

The standard enthalpy of formation of H2O can be determined using the enthalpy change of the reaction and stoichiometry.

Given that the reaction 2 H2 + O2 -> 2 H2O releases 483.6 kJ, we can divide this value by the stoichiometric coefficient of the product (2) to determine the enthalpy change per mole of H2O. Therefore, the standard enthalpy of formation of H2O is 241.8 kJ/mol.

Therefore, the correct answer is option D. +241.8 kJ.

answered
User Santthosh
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