Question

Calculate the enthalpy change of the reaction,
\(\text{H}_2\text{(g)}+\text{Cl}_2\text{(g)}\rightarrow 2\text{HCl(g)}\)
if bond energies (kJ mol\(^{-1}\)):
H–H = 436, Cl–Cl = 242, H–Cl = 431

• A. \(-184\) kJ/mol
• B. \(-246\) kJ/mol
• C. \(-242\) kJ/mol
• D. \(-431\) kJ/mol

Hint:

Enthalpy change equals bond energy of bonds broken minus bond energy of bonds formed.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Breaking bonds takes energy and forming bonds releases it. So the reaction enthalpy is found by comparing the two totals.

Step 2: Key Formula or Approach:
\[ \Delta H = \sum \text{(bond energies of bonds broken)} - \sum \text{(bond energies of bonds formed)} \]

Step 3: Detailed Explanation:
Bonds broken: one H-H (436) and one Cl-Cl (242). Total = \(436 + 242 = 678\) kJ/mol.
Bonds formed: two H-Cl bonds, since 2 moles of HCl form. Total = \(2 \times 431 = 862\) kJ/mol.
\[ \Delta H = 678 - 862 = -184\text{ kJ/mol} \]
The negative sign shows the reaction releases heat.

Step 4: Why the other options are wrong.
The values -242 and -431 are just single bond energies of Cl-Cl and H-Cl, not a difference between broken and formed totals. The value -246 does not follow from the data at all. Option (A) is the only value that follows the formula.

Final Answer:
Energy released on forming 2 H-Cl bonds exceeds the energy needed to break H-H and Cl-Cl by 184 kJ. \[ \boxed{-184\text{ kJ/mol}} \]