Question

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

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

Answer 1

Answer: (D)

The reaction between hydrogen and chlorine molecules to form hydrogen chloride. This reaction can be broken down into these simple steps:
$H _{2}( g )+ C l _{2}( g ) \rightarrow 2 H C l ( g )$

Step 1: Dissociation of Hydrogen and Chlorine molecules into their respective atoms
$H _{2}( g ) \rightarrow 2 H ( g )$
$Cl _{2}( g ) \rightarrow 2 Cl ( g )$

Step 2: Combination of these atoms to form hydrogen chloride $2 H ( g )+2 Cl ( g ) \rightarrow 2 H Cl ( g )$

Since in the first step, one-mole of each $H - H$ and $C l - C l$ bonds are broken, it is an endothermic process $\left(\Delta H _{1}>0\right)$ while in the second step two moles of $H Cl$ are being formed which releases energy making $\Delta H _{2}<0$

Now, $\Delta H _{1}=\Delta H _{ Cl - Cl }+\Delta H _{ H - H }$
$= 4 3 4 + 2 4 2$
$= 6 7 6 \,k J\, m o l ^{-1}$
Now, $\Delta H _{2} =- 2 \Delta H _{ H - Cl }$
$=862\, kJ \,mol ^{-1}$
Applying Hess Law, i.e, $\Delta H ^{\prime}=\Delta H _{1}+\Delta H _{2}$, we get
$\Delta H ^{\prime}=- 186 \,kJ\, mol ^{- 1 }$
This is the enthalpy of formation of two moles of $H C l$. Hence the enthalpy of formation of one mole of $H C l$ will be
$\Delta H =- 93 \,kJ \,mol ^{- 1 }$