JEE Challenger
More from Thermodynamics
ChemMedAdv 2012 • Paper 2

Calculate Bond Energy of Carbon Triple Bond in Acetylene

Using the provided information, compute the triple bond energy (in kJ mol−1\text{kJ mol}^{-1}) of the C≡C\text{C}\equiv\text{C} bond present in C2H2\text{C}_2\text{H}_2. Assume the bond dissociation energy of a C−H\text{C}-\text{H} bond is 350 kJ mol−1350\text{ kJ mol}^{-1}.

2C(s)+H2(g)⟶C2H2(g)ΔH=225 kJ mol−12C(s)⟶2C(g)ΔH=1410 kJ mol−1H2(g)⟶2H(g)ΔH=330 kJ mol−1\begin{array}{ll} 2\text{C(s)} + \text{H}_2(\text{g}) \longrightarrow \text{C}_2\text{H}_2(\text{g}) & \Delta H = 225\text{ kJ mol}^{-1} \\ 2\text{C(s)} \longrightarrow 2\text{C(g)} & \Delta H = 1410\text{ kJ mol}^{-1} \\ \text{H}_2(\text{g}) \longrightarrow 2\text{H(g)} & \Delta H = 330\text{ kJ mol}^{-1} \end{array}

Options

A

1165

B

837

C

865

D

815

Correct

Topics & Concepts

ThermodynamicsEnthalpy

Step-by-Step Solution

To determine the bond energy of the carbon-carbon triple bond (C≡C\text{C}\equiv\text{C}) in acetylene (C2H2\text{C}_2\text{H}_2), we can utilize the relation between the reaction enthalpy and the enthalpy changes of individual atomization and bond formation processes.

The given thermochemical equations are:

  1. 2C(s)+H2(g)⟶C2H2(g)ΔH1=225 kJ mol−12\text{C(s)} + \text{H}_2(\text{g}) \longrightarrow \text{C}_2\text{H}_2(\text{g}) \quad \Delta H_1 = 225\text{ kJ mol}^{-1}
  2. 2C(s)⟶2C(g)ΔH2=1410 kJ mol−12\text{C(s)} \longrightarrow 2\text{C(g)} \quad \Delta H_2 = 1410\text{ kJ mol}^{-1}
  3. H2(g)⟶2H(g)ΔH3=330 kJ mol−1\text{H}_2(\text{g}) \longrightarrow 2\text{H(g)} \quad \Delta H_3 = 330\text{ kJ mol}^{-1}

Also, the bond dissociation energy of a single C−H\text{C}-\text{H} bond is given as: B.E.(C−H)=350 kJ mol−1\text{B.E.}(\text{C}-\text{H}) = 350\text{ kJ mol}^{-1}

The enthalpy change for the formation of acetylene from its elemental states (Reaction 1) can be expressed in terms of the sublimation energy of carbon, the dissociation energy of H2\text{H}_2, and the bond energies present in C2H2\text{C}_2\text{H}_2:

ΔH1=ΔH2+ΔH3−[B.E.(C≡C)+2×B.E.(C−H)]\Delta H_1 = \Delta H_2 + \Delta H_3 - \left[ \text{B.E.}(\text{C}\equiv\text{C}) + 2 \times \text{B.E.}(\text{C}-\text{H}) \right]

Substituting the given values into the equation:

225=1410+330−[B.E.(C≡C)+2(350)]225 = 1410 + 330 - \left[ \text{B.E.}(\text{C}\equiv\text{C}) + 2(350) \right]

225=1740−B.E.(C≡C)−700225 = 1740 - \text{B.E.}(\text{C}\equiv\text{C}) - 700

225=1040−B.E.(C≡C)225 = 1040 - \text{B.E.}(\text{C}\equiv\text{C})

Rearranging the terms to solve for B.E.(C≡C)\text{B.E.}(\text{C}\equiv\text{C}):

B.E.(C≡C)=1040−225=815 kJ mol−1\text{B.E.}(\text{C}\equiv\text{C}) = 1040 - 225 = 815\text{ kJ mol}^{-1}

Therefore, the correct option is D.

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