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Thermodynamics question

2018 · Q104
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Thermodynamics question

2018 · Q104

NEETChemistryThermodynamicsMCQ+4 / −1
The bond dissociation energies of X2 , Y2 and XY are in the ratio of 1 : 0.5 : 1. Δ\DeltaΔH for the formation of XY is –200 kJ mol–1. The bond dissociation energy of X2 will be
  1. A
    800 kJ mol–1
  2. B
    200 kJ mol–1
  3. C
    400 kJ mol–1
  4. D
    100 kJ mol–1
View written solutionFree

Correct answer: A

Let bond dissociation energies of X2 , Y2 and XY are x kJ mol–1 , 0.5x kJ mol–1 and x kJ mol–1 respectively.

12{1 \over 2}21​X2 + 12{1 \over 2}21​Y2 →\to→ XY; Δ\Delta ΔH = -200 kJ mol-1

Δ\Delta ΔH = Σ(B.E)Reactant - Σ(B.E)Product

⇒\Rightarrow⇒ -200 = [12×x+12×0.5x]\left[ {{1 \over 2} \times x + {1 \over 2} \times 0.5x} \right][21​×x+21​×0.5x] - [1×x]\left[ {1 \times x} \right][1×x]

On solving, x = 800 kJ mol–1

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