Question

Calculate the enthalpy change for the following reaction, using given bond energy (kJ/mol)
(C-H = \(414\), H-O = \(463\), H-Cl = \(431\), C-Cl = \(326\) and C-O = \(335\))
\(\text{CH}_3\text{OH}_{(g)}+\text{HCl}_{(g)}\rightarrow \text{CH}_3\text{Cl}_{(g)}+\text{H}_2\text{O}_{(g)}\)

• A. \(-23\text{ kJmol}^{-1}\)
• B. \(-42\text{ kJmol}^{-1}\)
• C. \(-59\text{ kJmol}^{-1}\)
• D. \(-51\text{ kJmol}^{-1}\)

Hint:

$\Delta H=$ bonds broken $-$ bonds formed. Remember water has two O-H bonds.

Answer 1

The Correct Option is A

Step 1: Write the rule
\(\Delta H=\sum(\text{bond energies of reactants})-\sum(\text{bond energies of products})\).

Step 2: Bonds broken
In CH\(_3\)OH we break one C-O bond (\(335\)) and one O-H bond (\(463\)). In HCl we break H-Cl (\(431\)).
\[ 335+463+431=1229\text{ kJ} \]

Step 3: Bonds formed
In CH\(_3\)Cl we form one C-Cl bond (\(326\)). In H\(_2\)O we form two O-H bonds (\(2\times463=926\)).
\[ 326+926=1252\text{ kJ} \]

Step 4: Compute
\[ \Delta H=1229-1252=-23\text{ kJ mol}^{-1} \]
The C-H bonds are unchanged and cancel.

Final Answer:
\(\Delta H=-23\) kJ mol\(^{-1}\), option (A). \[ \boxed{\text{(A) } -23\text{ kJ mol}^{-1}} \]