The reaction of lithium (Li) with water (H₂O) is highly exothermic, meaning it releases heat. Here's how to calculate the energy released per mole of lithium:

Reaction:

2 Li(s) + 2 H₂O(l) → 2 LiOH(aq) + H₂(g)

Enthalpy Change (ΔH) Calculation:

You need to look up the standard enthalpies of formation (ΔHf°) for each compound involved in the reaction. These values are typically found in chemistry textbooks or online databases.

Here's a simplified example using hypothetical ΔHf° values (you'll need to look up the actual values):

* ΔHf° [LiOH(aq)] = -485 kJ/mol

* ΔHf° [H₂(g)] = 0 kJ/mol (by definition)

* ΔHf° [Li(s)] = 0 kJ/mol (by definition)

* ΔHf° [H₂O(l)] = -286 kJ/mol

Using Hess's Law:

ΔH = [Σ ΔHf° (products)] - [Σ ΔHf° (reactants)]

ΔH = [2 * (-485 kJ/mol) + 0] - [2 * 0 + 2 * (-286 kJ/mol)]

ΔH = -970 kJ/mol + 572 kJ/mol

ΔH = -398 kJ/mol

Important Note: This result is for the reaction of *two* moles of lithium. To get the energy per mole of lithium, divide by 2:

-398 kJ/mol / 2 = -199 kJ/mol

Therefore, the reaction of lithium with water releases approximately 199 kJ of energy per mole of lithium.

Safety:

Remember that this reaction is very vigorous and produces hydrogen gas, which is flammable. Always handle lithium and water with extreme caution and never perform this experiment without proper safety equipment and supervision.