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Chemistry Question on Thermodynamics

Bond dissociation enthalpy of H2,Cl2{H_2, Cl_2} and HClHCl are 434,242434, 242 and 431 kJmol−1{431 \, kJmol^{- 1}} respectively. Enthalpy of formation of HClHCl is

A

245 kJmol−1{245\, kJmol^{- 1}}

B

93 kJmol−1{93\, kJmol^{- 1}}

C

−245 kJmol−1{-245\, kJmol^{- 1}}

D

−93 kJmol−1{-93\, kJmol^{- 1}}

Answer

−93 kJmol−1{-93\, kJmol^{- 1}}

Explanation

Solution

The reaction between hydrogen and chlorine molecules to form hydrogen chloride. This reaction can be broken down into these simple steps:
H2(g)+Cl2(g)→2HCl(g)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
H2(g)→2H(g)H _{2}( g ) \rightarrow 2 H ( g )
Cl2(g)→2Cl(g)Cl _{2}( g ) \rightarrow 2 Cl ( g )

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

Since in the first step, one-mole of each H−HH - H and Cl−ClC l - C l bonds are broken, it is an
endothermic process (ΔH1>0)\left(\Delta H _{1}>0\right) while in the second step two moles of HClH Cl are being
formed which releases energy making ΔH2<0\Delta H _{2}<0

Now, ΔH1=ΔHCl−Cl+ΔHH−H\Delta H _{1}=\Delta H _{ Cl - Cl }+\Delta H _{ H - H }
=434+242= 4 3 4 + 2 4 2
=676kJmol−1= 6 7 6 k J m o l ^{-1}
Now, ΔH2=−2ΔHH−Cl\Delta H _{2} =- 2 \Delta H _{ H - Cl }
=862kJmol−1=862 kJ mol ^{-1}
Applying Hess Law, i.e, ΔH′=ΔH1+ΔH2\Delta H ^{\prime}=\Delta H _{1}+\Delta H _{2}, we get
ΔH′=−186kJmol−1\Delta H ^{\prime}=- 1 8 6 k J m o l ^{- 1 }
This is the enthalpy of formation of two moles of HClH C l. Hence the enthalpy of formation of one mole of HClH C l will be
ΔH=−93kJmol−1\Delta H =- 9 3 k J m o l ^{- 1 }