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JEE Advanced 2012Paper 2
Exam
JEE Advanced
Year
2012
Paper
Paper 2
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Advanced 2012 · Paper 2 · Chemistry · Thermodynamics (C)

Using the data provided, calculate the multiple bond energy (kJ...

Using the data provided, calculate the multiple bond energy (kJmol−1)\left(\mathrm{kJ} \mathrm{mol}^{-1}\right) of a C≡C\mathrm{C} \equiv \mathrm{C} bond in C2H2\mathrm{C}_{2} \mathrm{H}_{2}. That energy is (take the bond energy of a C−H\mathrm{C}-\mathrm{H} bond as 350 kJ mol−1350 \mathrm{~kJ} \mathrm{~mol}^{-1} ) 2C(s)+H2( g)⟶2 \mathrm{C}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow\mathrm{HC}=\mathrm{CH}(g);ΔH=225 kJ mol−12C(s)⟶2C(g);ΔH=1410 kJ mol−1H2( g)⟶2H(g);ΔH=330 kJ mol−1(\mathrm{g}) ; \Delta \mathrm{H}=225 \mathrm{~kJ} \mathrm{~mol}^{-1} 2 \mathrm{C}(\mathrm{s}) \longrightarrow 2 \mathrm{C}(\mathrm{g}) ; \Delta \mathrm{H}=1410 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}(\mathrm{g}) ; \Delta \mathrm{H}=330 \mathrm{~kJ} \mathrm{~mol}^{-1}

Options

  1. A1165
  2. B837
  3. C865
  4. D815

Correct answer

D. 815

Step-by-step solution

(i) 2C(s)+H2( g)⟶H−C≡C−H(g)2 \mathrm{C}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}(\mathrm{g}) ΔH=225 kJ mol−1\Delta \mathrm{H}=225 \mathrm{~kJ} \mathrm{~mol}^{-1} (ii) 2C(s)⟶2C(g)ΔH=1410 kJ mol−12 \mathrm{C}(\mathrm{s}) \longrightarrow 2 \mathrm{C}(\mathrm{g}) \Delta \mathrm{H}=1410 \mathrm{~kJ} \mathrm{~mol}^{-1} (iii) H2\mathrm{H}_{2} (g) ⟶2H\longrightarrow 2 \mathrm{H} (g) ΔH=330 kJ mol−1\Delta \mathrm{H}=330 \mathrm{~kJ} \mathrm{~mol}^{-1} From equation (i) : 225=[2×ΔHC(s)⟶C(g)+1×BEH−H]−[2×BEC−H+1×BEC≡C]225=[1410+1×330]−[2×350+1×BEC≡C]225=[1410+330]−[700+BEC≡C]225=1740−700−BEC≡CBEC≡C=815 kJ mol−1\begin{array}{l} 225=\left[2 \times \Delta \mathrm{H}_{\mathrm{C}(\mathrm{s}) \longrightarrow \mathrm{C}(\mathrm{g})}+1 \times \mathrm{BE}_{\mathrm{H}-\mathrm{H}}\right] \\ -\left[2 \times \mathrm{BE}_{\mathrm{C}-\mathrm{H}}+1 \times \mathrm{BE}_{\mathrm{C} \equiv \mathrm{C}}\right] \\ 225=[1410+1 \times 330]-\left[2 \times 350+1 \times \mathrm{BE}_{\mathrm{C} \equiv \mathrm{C}}\right] \\ 225=[1410+330]-\left[700+\mathrm{BE}_{\mathrm{C} \equiv \mathrm{C}}\right] \\ 225=1740-700-\mathrm{BE}_{\mathrm{C} \equiv \mathrm{C}} \\ \mathrm{BE}_{\mathrm{C} \equiv \mathrm{C}}=815 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{array}
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