Question

Bond dissociation enthalpy of ${H_2, Cl_2}$ and $HCl$ are $434, 242$ and ${431 \, kJmol^{- 1}}$ respectively. Enthalpy of formation of $HCl $ is

• A. ${245\, kJmol^{- 1}}$
• B. ${93\, kJmol^{- 1}}$
• C. ${-245\, kJmol^{- 1}}$
• D. ${-93\, kJmol^{- 1}}$

Answer 1

The Correct Option is D

To determine the enthalpy of formation of $HCl$, we need to understand the process of formation and apply Hess's Law. Given are the bond dissociation enthalpies for ${H_2}$, ${Cl_2}$, and ${HCl}$:

• $H_2$: \, 434 \, kJ/mol$
• $Cl_2$: \, 242 \, kJ/mol$
• $HCl$: \, 431 \, kJ/mol$

The reaction for the formation of $HCl$ is:

$\frac{1}{2} H_2 (g) + \frac{1}{2} Cl_2 (g) \rightarrow HCl (g)$

Using Hess’s Law, the enthalpy change for this reaction is the difference between the bond dissociation energies of the reactants and products:

\[\Delta H = \left(\frac{1}{2} \cdot 434 + \frac{1}{2} \cdot 242 - 431\right) \, kJ/mol\]

Calculating each term:

• Bond dissociation energy for $H_2$: \frac{1}{2} \cdot 434 = 217 \, kJ/mol$
• Bond dissociation energy for $Cl_2$: \frac{1}{2} \cdot 242 = 121 \, kJ/mol$
• Total dissociation energy for reactants: $217 + 121 = 338 \, kJ/mol$

Now, inserting these into the equation:

\[\Delta H = 338 - 431 = -93 \, kJ/mol\]

The enthalpy of formation of $HCl$ is thus $-93 \, kJ/mol$.

Therefore, the correct answer is:

${-93\, kJmol^{- 1}}$