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Thermodynamics question

2021 · 18 Mar · Shift 1 · Q18
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Thermodynamics question

2021 · 18 Mar · Shift 1 · Q18

JEE MainChemistryThermodynamicsNumerical+4 / −1
For the reaction C2H6C_2H_6C2​H6​ →\to→ C2H4C_2H_4C2​H4​ + H2H_2H2​ the reaction enthalpy Δ\DeltaΔ rH = ‾\underline{\hspace{2cm}}​ kJ mol −-− 1. (Round off to the Nearest Integer). [ Given : Bond enthalpies in kJ mol −-− 1 : C-C : 347, C = C : 611; C-H : 414, H-H : 436 ]
Numerical answer
View written solutionFree

Correct answer: 128

  1. Write the reaction

C2H6→C2H4+H2\mathrm{C_2H_6 \rightarrow C_2H_4 + H_2}C2​H6​→C2​H4​+H2​

We use average bond enthalpies:

  • C−C=347 kJ mol−1\mathrm{C-C} = 347\ \text{kJ mol}^{-1}C−C=347 kJ mol−1
  • C=C=611 kJ mol−1\mathrm{C=C} = 611\ \text{kJ mol}^{-1}C=C=611 kJ mol−1
  • C−H=414 kJ mol−1\mathrm{C-H} = 414\ \text{kJ mol}^{-1}C−H=414 kJ mol−1
  • H−H=436 kJ mol−1\mathrm{H-H} = 436\ \text{kJ mol}^{-1}H−H=436 kJ mol−1
  1. Use bond enthalpy relation

ΔrH=∑(bond enthalpies of bonds broken)−∑(bond enthalpies of bonds formed)\Delta_r H = \sum (\text{bond enthalpies of bonds broken}) - \sum (\text{bond enthalpies of bonds formed})Δr​H=∑(bond enthalpies of bonds broken)−∑(bond enthalpies of bonds formed)

  1. Compare reactant and products

Ethane, C2H6\mathrm{C_2H_6}C2​H6​, has:

  • 111 bond of C−C\mathrm{C-C}C−C
  • 666 bonds of C−H\mathrm{C-H}C−H

Ethene, C2H4\mathrm{C_2H_4}C2​H4​, has:

  • 111 bond of C=C\mathrm{C=C}C=C
  • 444 bonds of C−H\mathrm{C-H}C−H

Hydrogen, H2\mathrm{H_2}H2​, has:

  • 111 bond of H−H\mathrm{H-H}H−H

Common bonds on both sides cancel conceptually:

  • 444 C−H\mathrm{C-H}C−H bonds remain on both sides

So the effective change is:

  • Broken: 1 C−C1\,\mathrm{C-C}1C−C and 2 C−H2\,\mathrm{C-H}2C−H
  • Formed: 1 C=C1\,\mathrm{C=C}1C=C and 1 H−H1\,\mathrm{H-H}1H−H
  1. Calculate energy of bonds broken

Ebroken=1(347)+2(414)=347+828=1175 kJ mol−1E_{\text{broken}} = 1(347) + 2(414) = 347 + 828 = 1175\ \text{kJ mol}^{-1}Ebroken​=1(347)+2(414)=347+828=1175 kJ mol−1

  1. Calculate energy of bonds formed

Eformed=1(611)+1(436)=1047 kJ mol−1E_{\text{formed}} = 1(611) + 1(436) = 1047\ \text{kJ mol}^{-1}Eformed​=1(611)+1(436)=1047 kJ mol−1

  1. Find reaction enthalpy

ΔrH=1175−1047=128 kJ mol−1\Delta_r H = 1175 - 1047 = 128\ \text{kJ mol}^{-1}Δr​H=1175−1047=128 kJ mol−1

  1. Nearest integer

128\boxed{128}128​

So,

ΔrH=+128 kJ mol−1\boxed{\Delta_r H = +128\ \text{kJ mol}^{-1}}Δr​H=+128 kJ mol−1​

The reaction is endothermic.

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