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Thermodynamics question

2003 · Shift 0 · Q3
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Thermodynamics question

2003 · Shift 0 · Q3

JEE MainChemistryThermodynamicsMCQ+4 / −1
If at 298 K the bond energies of C - H, C - C, C = C and H - H bonds are respectively 414, 347, 615 and 435 kJ/mol, the value of enthalpy change for the reaction H2C=CH2H_2C=CH_2H2​C=CH2​(g) + H2H_2H2​(g) →\to→ H3C−CH3H_3C-CH_3H3​C−CH3​(g) at 298 K will be :
  1. A
    - 250 kJ
  2. B
    + 125 kJ
  3. C
    - 125 kJ
  4. D
    + 250 kJ
View written solutionFree

Correct answer: C

  1. Use bond energies to estimate enthalpy change

For a reaction,

ΔH=∑(bond energies of bonds broken)−∑(bond energies of bonds formed)\Delta H = \sum (\text{bond energies of bonds broken}) - \sum (\text{bond energies of bonds formed})ΔH=∑(bond energies of bonds broken)−∑(bond energies of bonds formed)
  1. Write the reaction
H2C=CH2(g)+H2(g)→H3C−CH3(g)\mathrm{H_2C{=}CH_2(g) + H_2(g) \to H_3C{-}CH_3(g)}H2​C=CH2​(g)+H2​(g)→H3​C−CH3​(g)

This is hydrogenation of ethene to ethane.

  1. Identify bonds broken and formed

Reactants:

  • In H2C=CH2\mathrm{H_2C{=}CH_2}H2​C=CH2​: one C=C\mathrm{C{=}C}C=C bond and four C−H\mathrm{C-H}C−H bonds
  • In H2\mathrm{H_2}H2​: one H−H\mathrm{H-H}H−H bond

Products:

  • In H3C−CH3\mathrm{H_3C-CH_3}H3​C−CH3​: one C−C\mathrm{C-C}C−C bond and six C−H\mathrm{C-H}C−H bonds

Now compare reactants and products:

  • 444 C−H\mathrm{C-H}C−H bonds are already present on both sides
  • Final product has 666 C−H\mathrm{C-H}C−H bonds, so effectively 2 new C−H\mathrm{C-H}C−H bonds are formed
  • The C=C\mathrm{C{=}C}C=C bond becomes a C−C\mathrm{C-C}C−C bond
  • One H−H\mathrm{H-H}H−H bond is broken

So effectively:

  • Bonds broken: 1×C=C1\times \mathrm{C{=}C}1×C=C and 1×H−H1\times \mathrm{H-H}1×H−H
  • Bonds formed: 1×C−C1\times \mathrm{C-C}1×C−C and 2×C−H2\times \mathrm{C-H}2×C−H
  1. Substitute given bond energies

Given:

  • C−H=414 kJ/mol\mathrm{C-H} = 414\ \text{kJ/mol}C−H=414 kJ/mol
  • C−C=347 kJ/mol\mathrm{C-C} = 347\ \text{kJ/mol}C−C=347 kJ/mol
  • C=C=615 kJ/mol\mathrm{C{=}C} = 615\ \text{kJ/mol}C=C=615 kJ/mol
  • H−H=435 kJ/mol\mathrm{H-H} = 435\ \text{kJ/mol}H−H=435 kJ/mol

Energy of bonds broken:

615+435=1050 kJ/mol615 + 435 = 1050\ \text{kJ/mol}615+435=1050 kJ/mol

Energy of bonds formed:

347+2(414)=347+828=1175 kJ/mol347 + 2(414) = 347 + 828 = 1175\ \text{kJ/mol}347+2(414)=347+828=1175 kJ/mol

Therefore,

ΔH=1050−1175=−125 kJ/mol\Delta H = 1050 - 1175 = -125\ \text{kJ/mol}ΔH=1050−1175=−125 kJ/mol
  1. Match with options
ΔH=−125 kJ/mol\boxed{\Delta H = -125\ \text{kJ/mol}}ΔH=−125 kJ/mol​

So the correct option is C.

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