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

Correct answer: D. -93 kJmol^ -1 aligned & 2[H]+2[Cl] 2 HCl, Δ H=-862 kJ mol^ -1 & H_ 2 2[ H], Δ H=434 kJ & Cl_ 2 2[Cl], Δ H=242 kJ & H_ 2 + Cl_ 2 2 HCl, Δ H=-186 kJ & Δ H_ 4 of HCl=(-186)/(2)=-93 kJ mol^ -1 aligned

NEET 2008

Exam: NEET
Year: 2008
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2008 · Chemistry · Thermodynamics (C)

Bond dissociation enthalpy of H_2, Cl_2 and HCl are 434,242 and 431...

Bond dissociation enthalpy of

\mathrm{H}_2, \mathrm{Cl}_2

and

\mathrm{HCl}

are 434,242 and

431 \mathrm{kJmol}^{-}

{ }^1

respectively. Enthalpy of formation of

\mathrm{HCl}

Options

A $245 \mathrm{kJmol}^{-1}$
B $93 \mathrm{kJmol}^{-1}$
C $-245 \mathrm{kJmol}^{-1}$
D $-93 \mathrm{kJmol}^{-1}$

Correct answer

D. $-93 \mathrm{kJmol}^{-1}$

Step-by-step solution

\begin{aligned}& 2[\mathrm{H}]+2[\mathrm{Cl}] \longrightarrow2 \mathrm{HCl}, \Delta \mathrm{H}=-862 \mathrm{ kJ} \mathrm{ mol}^{-1}\\ & \mathrm{H}_{2} \longrightarrow 2[ \mathrm{H}], \Delta \mathrm{H}=434 \mathrm{ kJ}\\ & \mathrm{Cl}_{2} \longrightarrow 2[\mathrm{Cl}], \Delta \mathrm{H}=242 \mathrm{ kJ}\\ & \mathrm{H}_{2}+ \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{HCl}, \Delta \mathrm{H}=-186 \mathrm{ kJ}\\ & \therefore \Delta \mathrm{H}_{4} \text{ of } \mathrm{HCl}=\frac{-186}{2}=-93 \mathrm{ kJ} \mathrm{ mol}^{-1} \end{aligned}

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