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Thermodynamics question

2025 · 23 Jan · Shift 2 · Q22
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  5. /2025 · 23 Jan · Shift 2 · Q22

Thermodynamics question

2025 · 23 Jan · Shift 2 · Q22

JEE MainChemistryThermodynamicsNumerical+4 / −1
The bond dissociation enthalpy of X2ΔHbond ∘\mathrm{X}_2 \Delta \mathrm{H}_{\text {bond }}^{\circ}X2​ΔHbond ∘​ calculated from the given data is ‾\underline{\hspace{2cm}}​kJmol−1\mathrm{kJ} \mathrm{mol}^{-1}kJmol−1. (Nearest integer) M+X−(s)→M+(g)+X−(g)ΔHlattice ∘=800 kJ mol−1M( s)→M( g)ΔHsub ∘=100 kJ mol−1M( g)→M+(g)+e−(g)ΔHi∘=500 kJ mol−1X( g)+e−(g)→X−(g)ΔHeg∘=−300 kJ mol−1M( s)+12X2( g)→M+X−(s)ΔHf∘=−400 kJ mol−1\begin{aligned} & \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\text {lattice }}^{\circ}=800 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \mathrm{M}(\mathrm{~s}) \rightarrow \mathrm{M}(\mathrm{~g}) \Delta \mathrm{H}_{\text {sub }}^{\circ}=100 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}\mathrm{M}(\mathrm{~g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{e}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{i}}^{\circ}=500 \mathrm{~kJ} \mathrm{~mol}^{-1}\mathrm{X}(\mathrm{~g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{eg}}^{\circ}=-300 \mathrm{~kJ} \mathrm{~mol}^{-1}\mathrm{M}(\mathrm{~s})+\frac{1}{2} \mathrm{X}_2(\mathrm{~g}) \rightarrow \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \Delta \mathrm{H}_f^{\circ}=-400 \mathrm{~kJ} \mathrm{~mol}^{-1}​M+X−(s)→M+(g)+X−(g)ΔHlattice ∘​=800 kJ mol−1M( s)→M( g)ΔHsub ∘​=100 kJ mol−1​M( g)→M+(g)+e−(g)ΔHi∘​=500 kJ mol−1X( g)+e−(g)→X−(g)ΔHeg∘​=−300 kJ mol−1M( s)+21​X2​( g)→M+X−(s)ΔHf∘​=−400 kJ mol−1[Given : M+X−\mathrm{M}^{+} \mathrm{X}^{-}M+X− is a pure ionic compound and X forms a diatomic molecule X2\mathrm{X}_2X2​ in gaseous state]
Numerical answer
View written solutionFree

Correct answer: 200

  1. Use Born–Haber cycle for the formation reaction
M(s)+12X2(g)→MX(s),ΔHf∘=−400 kJ mol−1\mathrm{M(s)}+\frac12\mathrm{X_2(g)}\rightarrow \mathrm{MX(s)}, \qquad \Delta H_f^\circ=-400\,\text{kJ mol}^{-1}M(s)+21​X2​(g)→MX(s),ΔHf∘​=−400kJ mol−1

Given steps:

  • Sublimation of metal:

    M(s)→M(g),ΔHsub∘=100\mathrm{M(s)}\rightarrow \mathrm{M(g)}, \qquad \Delta H_{sub}^\circ=100M(s)→M(g),ΔHsub∘​=100

  • Ionization of metal:

    M(g)→M+(g)+e−,ΔHi∘=500\mathrm{M(g)}\rightarrow \mathrm{M^+(g)}+e^-, \qquad \Delta H_i^\circ=500M(g)→M+(g)+e−,ΔHi∘​=500

  • Dissociation of halogen-like molecule:

    12X2(g)→X(g),ΔH=12D\frac12\mathrm{X_2(g)}\rightarrow \mathrm{X(g)}, \qquad \Delta H=\frac12 D21​X2​(g)→X(g),ΔH=21​D

    where D=ΔHbond∘D=\Delta H_{bond}^\circD=ΔHbond∘​ of X2\mathrm{X_2}X2​.

  • Electron gain:

    X(g)+e−→X−(g),ΔHeg∘=−300\mathrm{X(g)}+e^-\rightarrow \mathrm{X^-(g)}, \qquad \Delta H_{eg}^\circ=-300X(g)+e−→X−(g),ΔHeg∘​=−300

  • Lattice formation:

Given lattice dissociation:

MX(s)→M+(g)+X−(g),ΔHlattice∘=800\mathrm{MX(s)}\rightarrow \mathrm{M^+(g)}+\mathrm{X^-(g)}, \qquad \Delta H_{lattice}^\circ=800MX(s)→M+(g)+X−(g),ΔHlattice∘​=800

So lattice formation is the reverse process:

M+(g)+X−(g)→MX(s),ΔH=−800\mathrm{M^+(g)}+\mathrm{X^-(g)}\rightarrow \mathrm{MX(s)}, \qquad \Delta H=-800M+(g)+X−(g)→MX(s),ΔH=−800
  1. Apply Hess's law

Sum of all steps equals enthalpy of formation:

100+500+12D−300−800=−400100+500+\frac12D-300-800=-400100+500+21​D−300−800=−400
  1. Simplify
100+500−300−800+12D=−400100+500-300-800+\frac12D=-400100+500−300−800+21​D=−400 −500+12D=−400-500+\frac12D=-400−500+21​D=−400 12D=100\frac12D=10021​D=100 D=200 kJ mol−1D=200\,\text{kJ mol}^{-1}D=200kJ mol−1
  1. Final answer

The bond dissociation enthalpy of X2\mathrm{X_2}X2​ is

200 kJ mol−1\boxed{200\,\text{kJ mol}^{-1}}200kJ mol−1​
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