Calculate `Delta H` for the reaction : `2Li (s) + 2H_(2)O(l) rarr 2Li^(+) ( aq) + 2OH^(-) (aq) + H_(2)(g)`
Given that the standard enthalpies of formation of `Li^(+)(aq), OH^(-)(aq) ` and `H_(2)O(l)` are - 278.5, - 228.9 and - 285.8 kJ `mol^(-)` respectively ( all at 298 K )

Answer

To calculate the change in enthalpy (ΔH) for the reaction:

\[ 2 \text{Li (s)} + 2 \text{H}_2\text{O (l)} \rightarrow 2 \text{Li}^+ \text{(aq)} + 2 \text{OH}^- \text{(aq)} + \text{H}_2 \text{(g)} \]

we will use the standard enthalpies of formation (ΔH_f) of the products and reactants. The formula for calculating ΔH for the reaction is:

\[ \Delta H = \sum \Delta H_f \text{(products)} - \sum \Delta H_f \text{(reactants)} \]

### Step 1: Identify the standard enthalpies of formation for the products.

The products of the reaction are:
- \( \text{Li}^+ \text{(aq)} \)
- \( \text{OH}^- \text{(aq)} \)
- \( \text{H}_2 \text{(g)} \)

Given values:
- \( \Delta H_f \text{(Li}^+\text{(aq))} = -278.5 \, \text{kJ/mol} \)
- \( \Delta H_f \text{(OH}^-\text{(aq))} = -228.9 \, \text{kJ/mol} \)
- \( \Delta H_f \text{(H}_2\text{(g))} = 0 \, \text{kJ/mol} \) (since it is in its elemental form)

### Step 2: Calculate the total ΔH_f for the products.

For the products:
\[ \Delta H_f \text{(products)} = 2 \times \Delta H_f \text{(Li}^+\text{(aq))} + 2 \times \Delta H_f \text{(OH}^-\text{(aq))} + 1 \times \Delta H_f \text{(H}_2\text{(g))} \]
\[ = 2 \times (-278.5) + 2 \times (-228.9) + 0 \]
\[ = -557.0 - 457.8 + 0 = -1014.8 \, \text{kJ} \]

### Step 3: Identify the standard enthalpies of formation for the reactants.

The reactants of the reaction are:
- \( \text{Li (s)} \)
- \( \text{H}_2\text{O (l)} \)

Given values:
- \( \Delta H_f \text{(Li (s))} = 0 \, \text{kJ/mol} \) (since it is in its elemental form)
- \( \Delta H_f \text{(H}_2\text{O (l))} = -285.8 \, \text{kJ/mol} \)

### Step 4: Calculate the total ΔH_f for the reactants.

For the reactants:
\[ \Delta H_f \text{(reactants)} = 2 \times \Delta H_f \text{(Li (s))} + 2 \times \Delta H_f \text{(H}_2\text{O (l))} \]
\[ = 2 \times 0 + 2 \times (-285.8) \]
\[ = 0 - 571.6 = -571.6 \, \text{kJ} \]

### Step 5: Calculate ΔH for the reaction.

Now substituting the values into the ΔH equation:
\[ \Delta H = \Delta H_f \text{(products)} - \Delta H_f \text{(reactants)} \]
\[ = -1014.8 - (-571.6) \]
\[ = -1014.8 + 571.6 = -443.2 \, \text{kJ} \]

### Final Answer:
The change in enthalpy (ΔH) for the reaction is:

\[ \Delta H = -443.2 \, \text{kJ} \]