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Thermodynamics question

2023 · 31 Jan · Shift 1 · Q18
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Thermodynamics question

2023 · 31 Jan · Shift 1 · Q18

JEE MainChemistryThermodynamicsNumerical+4 / −1
The enthalpy change for the conversion of 12Cl2( g)\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{~g})21​Cl2​( g) to Cl−\mathrm{Cl}^{-}Cl−(aq) is (−-−) ‾kJmol−1\underline{\hspace{2cm}}\mathrm{kJ} \mathrm{mol}^{-1}​kJmol−1(Nearest integer) Given : ΔdisHCl2( g)θ⊖=240 kJ mol−1,ΔegHCl(g)⊖=−350 kJ mol−1\Delta_{\mathrm{dis}} \mathrm{H}_{\mathrm{Cl}_{2(\mathrm{~g})}^{\theta}}^{\ominus}=240 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\mathrm{eg}} \mathrm{H}_{\mathrm{Cl_{(g)}}}^{\ominus}=-350 \mathrm{~kJ} \mathrm{~mol}^{-1}Δdis​HCl2( g)θ​⊖​=240 kJ mol−1,Δeg​HCl(g)​⊖​=−350 kJ mol−1, ΔhydHCl(g)−Θ=−380{\mathrm{\Delta _{hyd}}H_{Cl_{(g)}^ - }^\Theta = - 380}Δhyd​HCl(g)−​Θ​=−380 kJ mol−1\mathrm{kJ~mol^{-1}}kJ mol−1
Numerical answer
View written solutionFree

Correct answer: 610

  1. We need the enthalpy change for

12 Cl2(g)→Cl−(aq)\frac{1}{2}\,\mathrm{Cl_2(g)} \rightarrow \mathrm{Cl^-(aq)}21​Cl2​(g)→Cl−(aq)

This conversion can be broken into three steps:

  • dissociation of 12Cl2(g)\frac12\mathrm{Cl_2(g)}21​Cl2​(g) to Cl(g)\mathrm{Cl(g)}Cl(g)
  • addition of electron to form Cl−(g)\mathrm{Cl^-(g)}Cl−(g)
  • hydration of Cl−(g)\mathrm{Cl^-(g)}Cl−(g) to Cl−(aq)\mathrm{Cl^-(aq)}Cl−(aq)
  1. Step 1: Dissociation

Given:

ΔHdis∘(Cl2(g))=240 kJ mol−1\Delta H_{\text{dis}}^\circ(\mathrm{Cl_2(g)}) = 240\ \mathrm{kJ\ mol^{-1}}ΔHdis∘​(Cl2​(g))=240 kJ mol−1

So for

Cl2(g)→2Cl(g)\mathrm{Cl_2(g)} \rightarrow 2\mathrm{Cl(g)}Cl2​(g)→2Cl(g)

the enthalpy change is +240 kJ mol−1+240\ \mathrm{kJ\ mol^{-1}}+240 kJ mol−1.

Hence for

12Cl2(g)→Cl(g)\frac12\mathrm{Cl_2(g)} \rightarrow \mathrm{Cl(g)}21​Cl2​(g)→Cl(g)

ΔH1=2402=+120 kJ mol−1\Delta H_1 = \frac{240}{2} = +120\ \mathrm{kJ\ mol^{-1}}ΔH1​=2240​=+120 kJ mol−1

  1. Step 2: Electron gain

Given:

ΔHeg∘(Cl(g))=−350 kJ mol−1\Delta H_{eg}^\circ(\mathrm{Cl(g)}) = -350\ \mathrm{kJ\ mol^{-1}}ΔHeg∘​(Cl(g))=−350 kJ mol−1

So

Cl(g)+e−→Cl−(g)\mathrm{Cl(g)} + e^- \rightarrow \mathrm{Cl^-(g)}Cl(g)+e−→Cl−(g)

has

ΔH2=−350 kJ mol−1\Delta H_2 = -350\ \mathrm{kJ\ mol^{-1}}ΔH2​=−350 kJ mol−1

  1. Step 3: Hydration of chloride ion

Given:

ΔHhyd∘(Cl−(g))=−380 kJ mol−1\Delta H_{hyd}^\circ(\mathrm{Cl^-(g)}) = -380\ \mathrm{kJ\ mol^{-1}}ΔHhyd∘​(Cl−(g))=−380 kJ mol−1

So

Cl−(g)→Cl−(aq)\mathrm{Cl^-(g)} \rightarrow \mathrm{Cl^-(aq)}Cl−(g)→Cl−(aq)

has

ΔH3=−380 kJ mol−1\Delta H_3 = -380\ \mathrm{kJ\ mol^{-1}}ΔH3​=−380 kJ mol−1

  1. Total enthalpy change

Adding all steps:

ΔH=ΔH1+ΔH2+ΔH3\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3ΔH=ΔH1​+ΔH2​+ΔH3​

ΔH=120−350−380\Delta H = 120 - 350 - 380ΔH=120−350−380

ΔH=−610 kJ mol−1\Delta H = -610\ \mathrm{kJ\ mol^{-1}}ΔH=−610 kJ mol−1

  1. The question is written as:

(−) ‾ kJ mol−1(-)\ \underline{\hspace{2cm}}\ \mathrm{kJ\ mol^{-1}}(−) ​ kJ mol−1

So the blank should contain the magnitude:

610610610

Therefore, the required integer is:

610\boxed{610}610​

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