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NEET 2008
Exam
NEET
Year
2008
Subject
Chemistry
Chapter
Chemical Equilibrium

NEET 2008 · Chemistry · Chemical Equilibrium

Bond dissociation enthalpy of H_2, Cl_2 and HCl Enthalpy of...

Bond dissociation enthalpy of H2,Cl2\mathrm{H}_2, \mathrm{Cl}_2 and HCl\mathrm{HCl} Enthalpy of formation of HCl\mathrm{HCl} is

Options

  1. A93 kJ mol−193 \mathrm{~kJ} \mathrm{~mol}^{-1}
  2. B−245 kJ mol−1-245 \mathrm{~kJ} \mathrm{~mol}^{-1}
  3. C−93 kJ mol−1-93 \mathrm{~kJ} \mathrm{~mol}^{-1}
  4. D245 kJ mol−1245 \mathrm{~kJ} \mathrm{~mol}^{-1}

Correct answer

C. −93 kJ mol−1-93 \mathrm{~kJ} \mathrm{~mol}^{-1}

Step-by-step solution

Key Idea: ΔHreaction =Σ\Delta H_{\text{reaction }}=\Sigma Bond energy of reactant −Σ-\Sigma Bond energy of product Here, ΔHH−H=434kJmol−1ΔHCl−Cl=242kJmol−1ΔHH−Cl=431kJmol−1∵12H2+12Cl2⟶HCl\begin{array}{ll} \text{Here, } & \Delta \mathrm{H}_{\mathrm{H}-\mathrm{H}}=434 \mathrm{ kJ} \mathrm{ mol}^{-1}\\ & \Delta \mathrm{H}_{\mathrm{Cl}-\mathrm{Cl}}=242 \mathrm{ kJ} \mathrm{ mol}^{-1} \\ & \Delta \mathrm{H}_{\mathrm{H}-\mathrm{Cl}}=431 \mathrm{ kJ} \mathrm{ mol}^{-1}\\ \because & \frac{1}{2}\mathrm{H}_2+\frac{1}{2} \mathrm{Cl}_2 \longrightarrow \mathrm{HCl} \end{array}ΔHreaction =12ΔHH−H+12ΔHCl−Cl−ΔHH−Cl=12×434+12×242−431=217+121−431=−93kJmol−1\begin{aligned} \Delta H_{\text{reaction }} &=\frac{1}{2} \Delta H_{\mathrm{H}-\mathrm{H}}+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{Cl}-\mathrm{Cl}}-\Delta \mathrm{H}_{\mathrm{H}-\mathrm{Cl}} \\ &=\frac{1}{2} \times 434+\frac{1}{2} \times 242-431 \\ & =217+121-431 \\ & =-93 \mathrm{ kJ}\mathrm{ mol}^{-1} \end{aligned}$
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