Bond energies of `(H - H), (O = O) and (O - H)` are 105, 120 and 220 kcal/mol respectively, then magnitude of `Delta H` in the reaction in kcal is :
`2H_(2)(g) + O_(2)(g) rarr 2H_(2)O(l)`

Explanation

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

\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \]

we will use the bond energies provided:

- Bond energy of \( H-H \) = 105 kcal/mol
- Bond energy of \( O=O \) = 120 kcal/mol
- Bond energy of \( O-H \) = 220 kcal/mol

### Step 1: Calculate the energy required to break the bonds in the reactants.

1. **Breaking the bonds in \( 2H_2 \)**:
- Each \( H_2 \) molecule has 1 \( H-H \) bond.
- For 2 \( H_2 \) molecules, we have:
\[ \text{Energy required} = 2 \times 105 \text{ kcal} = 210 \text{ kcal} \]

2. **Breaking the bonds in \( O_2 \)**:
- Each \( O_2 \) molecule has 1 \( O=O \) bond.
- For 1 \( O_2 \) molecule, we have:
\[ \text{Energy required} = 120 \text{ kcal} \]

### Step 2: Total energy required to break all bonds in the reactants.

\[ \text{Total energy to break bonds} = 210 \text{ kcal} + 120 \text{ kcal} = 330 \text{ kcal} \]

### Step 3: Calculate the energy released when new bonds are formed in the products.

1. **Forming the bonds in \( 2H_2O \)**:
- Each \( H_2O \) molecule has 2 \( O-H \) bonds.
- For 2 \( H_2O \) molecules, we have:
\[ \text{Total O-H bonds} = 2 \times 2 = 4 \text{ O-H bonds} \]
- The energy released when forming 4 \( O-H \) bonds:
\[ \text{Energy released} = 4 \times 220 \text{ kcal} = 880 \text{ kcal} \]

### Step 4: Calculate the overall change in enthalpy (ΔH).

\[ \Delta H = \text{Energy required to break bonds} - \text{Energy released from forming bonds} \]
\[ \Delta H = 330 \text{ kcal} - 880 \text{ kcal} = -550 \text{ kcal} \]

### Conclusion

The magnitude of \( \Delta H \) for the reaction is:

\[ \Delta H = -550 \text{ kcal} \]

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