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Thermodynamics question

2011 · Q114
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Thermodynamics question

2011 · Q114

NEETChemistryThermodynamicsMCQ+4 / −1
Enthalpy change for the reaction,
4H(g)  →\to→  2H2(g) is −-−869.6 kJ
The dissociation energy of H −-− H bond is
  1. A
    434.8 kJ
  2. B
    −-− 869.6 kJ
  3. C
    + 434.8 kJ
  4. D
    + 217.4 kJ
View written solutionFree

Correct answer: C

4H(g) → 2H2(g),    ∆H = – 869.6 kJ

Reverse the above equation

2H2(g) → 4H(g),    ∆H = + 869.6 kJ

Divide the above equation by 2,

H2(g) → 2H(g), ΔH=869.62\Delta H = {{869.6} \over 2}ΔH=2869.6​ kJ = 434.8 kJ

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