Using the data provided, calculate the multiple bond energy $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$ of a $\mathrm{C} \equiv \mathrm{C}$ bond in $\mathrm{C}_{2} \mathrm{H}_{2}$. That energy is (take the bond energy of a C-H bond as $350 \mathrm{~kJ} \mathrm{~mol}^{-1}$.) \[ \begin{aligned} 2 \mathrm{C}(\mathrm{~s})+\mathrm{H}_{2}(\mathrm{~g}) & \longrightarrow \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) & & \Delta H=225 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ 2 \mathrm{C}(\mathrm{~s}) & \longrightarrow 2 \mathrm{C}(\mathrm{~g}) & & \Delta H=1410 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{H}_{2}(\mathrm{~g}) & \longrightarrow 2 \mathrm{H}(\mathrm{~g}) & & \Delta H=330 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} \]

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Chemical Thermodynamics

Using the data provided, calculate the multiple bond energy

(kJ mol^{- 1})

of a

C \equiv C

bond in

C_{2} H_{2}

. That energy is (take the bond energy of a C-H bond as

350 kJ mol^{- 1}

2 C (s) + H_{2} (g) 2 C (s) H_{2} (g) ⟶ C_{2} H_{2} (g) ⟶ 2 C (g) ⟶ 2 H (g) Δ H = 225 kJ mol^{- 1} Δ H = 1410 kJ mol^{- 1} Δ H = 330 kJ mol^{- 1}

(A)1165
(B)837
(C)865
(D)815✓ Correct

Answer: (D)

Solution

Using Hess's law: ΔH(formation) = [ΔH(atomization of C) + ΔH(atomization of H₂)] - [2×BE(C-H) + BE(C≡C)]. Substituting: 225 = [1410 + 330] - [2×350 + BE(C≡C)]. Solving: 225 = 1740 - 700 - BE(C≡C), so BE(C≡C) = 1040 - 225 = 815 kJ/mol. The answer is (D).

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Single correct option

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