Calculate the enthalpy for the following reaction using the given bond energies (kJ/mol) :
`(C-H=414, H-O =463, H-Cl = 431, C-Cl=326, C-O =335)`
`CH_(3)-OH(g)+HClrarrCH_(3)-Cl(g)+H_(2)O(g)`

Options

A. `-23` kJ/mol
B. `-42` kJ/mol
C. `-59` kJ/mol
D. `-511` kJ/mol

Explanation

To calculate the enthalpy change for the reaction:

\[ \text{CH}_3\text{OH(g)} + \text{HCl} \rightarrow \text{CH}_3\text{Cl(g)} + \text{H}_2\text{O(g)} \]

we will use the bond energies provided. The bond energies are as follows:

- C-H = 414 kJ/mol
- H-O = 463 kJ/mol
- H-Cl = 431 kJ/mol
- C-Cl = 326 kJ/mol
- C-O = 335 kJ/mol

### Step 1: Identify the bonds broken in the reactants
In the reactants, we have:
- In CH₃OH:
- 3 C-H bonds
- 1 O-H bond
- 1 C-O bond
- In HCl:
- 1 H-Cl bond

So, the total number of bonds broken is:
- 3 C-H bonds
- 1 O-H bond
- 1 C-O bond
- 1 H-Cl bond

### Step 2: Calculate the total energy of bonds broken
Using the bond energies:
- Energy for breaking 3 C-H bonds = \(3 \times 414 = 1242 \, \text{kJ}\)
- Energy for breaking 1 O-H bond = \(1 \times 463 = 463 \, \text{kJ}\)
- Energy for breaking 1 C-O bond = \(1 \times 335 = 335 \, \text{kJ}\)
- Energy for breaking 1 H-Cl bond = \(1 \times 431 = 431 \, \text{kJ}\)

Now, sum these energies:
\[ \text{Total energy of bonds broken} = 1242 + 463 + 335 + 431 = 2471 \, \text{kJ} \]

### Step 3: Identify the bonds formed in the products
In the products, we have:
- In CH₃Cl:
- 3 C-H bonds
- 1 C-Cl bond
- In H₂O:
- 2 O-H bonds

So, the total number of bonds formed is:
- 3 C-H bonds
- 1 C-Cl bond
- 2 O-H bonds

### Step 4: Calculate the total energy of bonds formed
Using the bond energies:
- Energy for forming 3 C-H bonds = \(3 \times 414 = 1242 \, \text{kJ}\)
- Energy for forming 1 C-Cl bond = \(1 \times 326 = 326 \, \text{kJ}\)
- Energy for forming 2 O-H bonds = \(2 \times 463 = 926 \, \text{kJ}\)

Now, sum these energies:
\[ \text{Total energy of bonds formed} = 1242 + 326 + 926 = 2494 \, \text{kJ} \]

### Step 5: Calculate the enthalpy change (ΔH)
The enthalpy change for the reaction can be calculated using the formula:
\[ \Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed} \]
Substituting the values:
\[ \Delta H = 2471 - 2494 = -23 \, \text{kJ} \]

### Final Answer
The enthalpy change for the reaction is:
\[ \Delta H = -23 \, \text{kJ/mol} \]

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