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Thermodynamics question

2008 · Shift 0 · Q10
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  5. /2008 · Shift 0 · Q10

Thermodynamics question

2008 · Shift 0 · Q10

JEE MainChemistryThermodynamicsMCQ+4 / −1
Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below: 12Cl2(g){1 \over 2}C{l_2}(g)21​Cl2​(g) →12ΔdissHΘ\xrightarrow{{1 \over 2}{\Delta _{diss}}{H^\Theta }}21​Δdiss​HΘ​ Cl(g)→ΔegHΘCl−(g)→ΔHydHΘCl−(aq)Cl(g)\xrightarrow{{\Delta _{eg}}{H^\Theta }} C{l^ - }(g)\xrightarrow{{\Delta _{Hyd}}{H^\Theta }} C{l^ - }(aq)Cl(g)Δeg​HΘ​Cl−(g)ΔHyd​HΘ​Cl−(aq)(Using the data, ΔdissHCl2Θ{\Delta _{diss}}H_{C{l_2}}^\ThetaΔdiss​HCl2​Θ​= 240 kJ/mol, ΔegHClΘ{\Delta _{eg}}H_{Cl}^\ThetaΔeg​HClΘ​= -349 kJ/mol, ΔhydHCl−Θ{\Delta _{hyd}}H_{C{l^ - }}^\ThetaΔhyd​HCl−Θ​ = - 381 kJ/mol) will be :
  1. A
    +152 kJ mol−1
  2. B
    −610 kJ mol−1
  3. C
    −850 kJ mol−1
  4. D
    +120 kJ mol−1
View written solutionFree

Correct answer: B

  1. Write the given stepwise process

We need the enthalpy change for:

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

This is broken into three steps:

12Cl2(g)→12ΔdissH∘Cl(g)\frac{1}{2}Cl_2(g) \xrightarrow{\frac{1}{2}\Delta_{\text{diss}}H^\circ} Cl(g)21​Cl2​(g)21​Δdiss​H∘​Cl(g)

Cl(g)+e−→ΔegH∘Cl−(g)Cl(g) + e^- \xrightarrow{\Delta_{eg}H^\circ} Cl^-(g)Cl(g)+e−Δeg​H∘​Cl−(g)

Cl−(g)→ΔhydH∘Cl−(aq)Cl^-(g) \xrightarrow{\Delta_{hyd}H^\circ} Cl^-(aq)Cl−(g)Δhyd​H∘​Cl−(aq)

  1. Use the given data

  • Bond dissociation enthalpy of Cl2Cl_2Cl2​: ΔdissHCl2∘=240 kJ mol−1\Delta_{\text{diss}}H^\circ_{Cl_2} = 240\ \text{kJ mol}^{-1}Δdiss​HCl2​∘​=240 kJ mol−1 Therefore, 12ΔdissH∘=2402=120 kJ mol−1\frac{1}{2}\Delta_{\text{diss}}H^\circ = \frac{240}{2} = 120\ \text{kJ mol}^{-1}21​Δdiss​H∘=2240​=120 kJ mol−1

  • Electron gain enthalpy: ΔegHCl∘=−349 kJ mol−1\Delta_{eg}H^\circ_{Cl} = -349\ \text{kJ mol}^{-1}Δeg​HCl∘​=−349 kJ mol−1

  • Hydration enthalpy: ΔhydHCl−∘=−381 kJ mol−1\Delta_{hyd}H^\circ_{Cl^-} = -381\ \text{kJ mol}^{-1}Δhyd​HCl−∘​=−381 kJ mol−1

  1. Add all enthalpy changes

So total enthalpy change is:

ΔH∘=120+(−349)+(−381)\Delta H^\circ = 120 + (-349) + (-381)ΔH∘=120+(−349)+(−381)

ΔH∘=120−349−381\Delta H^\circ = 120 - 349 - 381ΔH∘=120−349−381

ΔH∘=−610 kJ mol−1\Delta H^\circ = -610\ \text{kJ mol}^{-1}ΔH∘=−610 kJ mol−1

  1. Match with the options

The correct option is:

B: −610 kJ mol−1\boxed{\text{B: } -610\ \text{kJ mol}^{-1}}B: −610 kJ mol−1​

  1. Comparison with stored answer

Stored correct answer is B, which matches our derived result.

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