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Thermodynamics question

2001 · Q115
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Thermodynamics question

2001 · Q115

NEETChemistryThermodynamicsMCQ+4 / −1
Change in enthalpy for reaction,

2H2O2(l) →\to→ 2H2O(l) + O2(g)

if heat of formation of H2O2(l) and H2O(l) are −-−188 and - 286 kJ/mol respectively, is
  1. A
    −-− 196 kJ/mol
  2. B
    + 196 kJ/mol
  3. C
    +948 kJ/mol
  4. D
    −-− 948 kJ/mol
View written solutionFree

Correct answer: A

2H2O2(l) →\to→ 2H2O(l) + O2(g), Δ\Delta ΔHr = ?

H2(g)

  • O2(g) → H2O2(l), Δ\Delta ΔH = – 188 kJ mol ...(1)

    H2(g)
  • O2(g) → H2O(l), Δ\Delta ΔH = –286 kJ/mol ....(2)

    (1) - (2)

    H2O2(l) → H2O(l)

12{1 \over 2}21​O2(g)

Δ\Delta ΔH = –286 – (– 188) = –98 kJ mol–1

Multiplying this equation by 2 we get the required equation

2H2O2(l) → 2H2O(l)

  • O2(g)

    Δ\Delta ΔH = 2 (– 98) kJ/mol = – 196 kJ/mol
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