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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Chemistry
Chapter
Thermodynamics (C)

BITSAT 2013 · Chemistry · Thermodynamics (C)

For the dissociation reaction, H_ 2 ( g) arrow 2 H(g) Δ H=162 Kcal...

For the dissociation reaction, H2(g)→2H(g)ΔH=162Kcal\mathrm{H}_{2}(\mathrm{ g}) \rightarrow 2 \mathrm{H}(\mathrm{g}) \quad \Delta \mathrm{H}=162 \mathrm{Kcal} heat of atomisation of H\mathrm{H} is

Options

  1. A81Kcal81 \mathrm{Kcal}
  2. B162Kcal162 \mathrm{Kcal}
  3. C208Kcal208 \mathrm{Kcal}
  4. D218Kcal218 \mathrm{Kcal}

Correct answer

A. 81Kcal81 \mathrm{Kcal}

Step-by-step solution

ΔH=ΔH(product) −ΔH(reactant) 162=2×ΔHH−ΔHH2ΔHH=1622(∵ΔHH2=0)ΔHH=81\Delta \mathrm{H}=\Delta \mathrm{H}_{\text{(product) }}-\Delta \mathrm{H}_{\text{(reactant) }}162=2 \times \Delta \mathrm{H}_{\mathrm{H}}-\Delta \mathrm{H}_{\mathrm{H}_{2}}\Delta \mathrm{H}_{\mathrm{H}}=\frac{162}{2} \quad\left(\because \Delta \mathrm{H}_{\mathrm{H}_{2}}=0\right)\Delta \mathrm{H}_{\mathrm{H}}=81Kcal\mathrm{Kcal}
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