What is the standard enthalpy of reaction for the following reaction:
H2 + 1/2 O2 --> H2O(g)
http://chemistry.about.com/od/thermodynamics/a/Heats-Of-Formation.htm
To find the standard enthalpy of reaction for the equation H2 + 1/2 O2 --> H2O(g), we need to subtract the enthalpy of the reactants from the enthalpy of the products.
First, we need to know the standard enthalpy of formation of each molecule involved in the reaction. The standard enthalpy of formation (ΔHf°) is the enthalpy change that occurs when one mole of a compound is formed from its constituent elements in their standard states at a given temperature and pressure.
In this case, we need the standard enthalpy of formation for H2, O2, and H2O(g).
The standard enthalpy of formation of H2 is 0 kJ/mol because it is in its standard state, meaning it already exists in its elemental form.
The standard enthalpy of formation of O2 is also 0 kJ/mol because it is also in its standard state.
The standard enthalpy of formation of H2O(g), however, is -241.83 kJ/mol. This value represents the enthalpy change when one mole of H2O(g) is formed from its constituent elements (H2 and O2) in their standard states.
Now, we can calculate the standard enthalpy of reaction using the formula:
ΔH° = Σ(n * ΔHf° products) - Σ(m * ΔHf° reactants)
In this case, we have one mole of H2O(g) as the product and one mole of H2 and 1/2 mole of O2 as the reactants.
Plugging in the values:
ΔH° = (1 * ΔHf° H2O(g)) - (1 * ΔHf° H2 + 1/2 * ΔHf° O2)
= (1 * -241.83 kJ/mol) - (1 * 0 kJ/mol + 1/2 * 0 kJ/mol)
= -241.83 kJ/mol
Therefore, the standard enthalpy of reaction for H2 + 1/2 O2 --> H2O(g) is -241.83 kJ/mol.