To determine the enthalpy change ([tex]\(\Delta H\)[/tex]) for the given reaction:
[tex]\[2 H_2 O_2 \rightarrow 2 H_2 O + O_2\][/tex]
we will follow these steps:
1. Identify the bonds broken in the reactants.
2. Identify the bonds formed in the products.
3. Calculate the total energy required to break the bonds in the reactants.
4. Calculate the total energy released when new bonds are formed in the products.
5. Subtract the energy required to break the bonds from the energy released to determine the enthalpy change ([tex]\(\Delta H\)[/tex]).
### Step 1: Bonds Broken in the Reactants
The structure of hydrogen peroxide ([tex]\(H_2O_2\)[/tex]) is:
[tex]\[ H - O - O - H \][/tex]
For 2 [tex]\(H_2O_2\)[/tex] molecules, the bonds broken are:
- 4 [tex]\(O-H\)[/tex] bonds
- 2 [tex]\(O-O\)[/tex] bonds
### Step 2: Bonds Formed in the Products
The products are 2 [tex]\(H_2O\)[/tex] and 1 [tex]\(O_2\)[/tex].
The structure of water ([tex]\(H_2O\)[/tex]) is:
[tex]\[ H - O - H \][/tex]
The structure of oxygen ([tex]\(O_2\)[/tex]) is:
[tex]\[ O = O \][/tex]
For the products, the bonds formed are:
- 4 [tex]\(O-H\)[/tex] bonds (from 2 [tex]\(H_2O\)[/tex] molecules)
- 1 [tex]\(O=O\)[/tex] bond (from [tex]\(O_2\)[/tex])
### Step 3: Total Energy for Bonds Broken in the Reactants
Using the given bond energies,
[tex]\[
\begin{aligned}
& \text{Bond energy for } O-H = 459 \text{ kJ/mol}\\
& \text{Bond energy for } O-O = 142 \text{ kJ/mol}\\
\end{aligned}
\][/tex]
Total energy for bonds broken:
[tex]\[
= 4 \times (459 \text{ kJ}) + 2 \times (142 \text{ kJ})
= 1836 \text{ kJ} + 284 \text{ kJ}
= 2120 \text{ kJ}
\][/tex]
### Step 4: Total Energy for Bonds Formed in the Products
Using the given bond energies,
[tex]\[
\begin{aligned}
& \text{Bond energy for } O-H = 459 \text{ kJ/mol}\\
& \text{Bond energy for } O=O = 494 \text{ kJ/mol}\\
\end{aligned}
\][/tex]
Total energy for bonds formed:
[tex]\[
= 4 \times (459 \text{ kJ}) + 1 \times (494 \text{ kJ})
= 1836 \text{ kJ} + 494 \text{ kJ}
= 2330 \text{ kJ}
\][/tex]
### Step 5: Calculate the Enthalpy Change ([tex]\(\Delta H\)[/tex])
[tex]\[
\Delta H = \text{Energy required to break bonds} - \text{Energy released in forming bonds}
\][/tex]
[tex]\[
\Delta H = 2120 \text{ kJ} - 2330 \text{ kJ} = -210 \text{ kJ}
\][/tex]
Therefore, the enthalpy change for the reaction is [tex]\(-210 \text{ kJ}\)[/tex].
The correct answer is:
[tex]\[ \boxed{-210 \text{ kJ}}\][/tex]
So, the correct option is:
[tex]\[ B. -210 \text{ kJ} \][/tex]
