What is the heat of hydrogenation of ethene if the bond energies of C-C , C=C , H-H , and C-H are 350,600,440 and 410kJ/mole respectively?

Question

What is the heat of hydrogenation of ethene if the bond energies of C-C , C=C , H-H , and C-H are 350,600,440 and 410kJ/mole respectively? (A) -130kJmol- 1 (B) -260kJmol- 1 (C) -400kJmol- 1 (D) -450kJmol- 1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-roe51nma8uxp2qoc.jpg" /> Δ H= Sigma Bond energy of reactants - Sigma Bond energy of products Δ H=[4 Δ HC - H + Δ HC = C + Δ HH - H] -[Δ HC - C + 6 Δ HC - H] =[.4× 410+600+440].-[.350+2460]. =2680-2810=-130kJmol- 1

Q. What is the heat of hydrogenation of ethene if the bond energies of $C - C$ , $C = C$ , $H - H$ , and $C - H$ are $350, 600, 440$ and $410 k J / m o l e$ respectively?

$- 130 k J m o l^{- 1}$

$- 260 k J m o l^{- 1}$

$- 400 k J m o l^{- 1}$

$- 450 k J m o l^{- 1}$

$Δ H = Σ$ Bond energy of reactants
$- Σ$ Bond energy of products
$Δ H = [4Δ H_{C - H} + Δ H_{C = C} + Δ H_{H - H}]$ $- [Δ H_{C - C} + 6Δ H_{C - H}]$
$= [4 \times 410 + 600 + 440] - [350 + 2460]$
$= 2680 - 2810 = - 130 k J m o l^{- 1}$

Q. What is the heat of hydrogenation of ethene if the bond energies of C−C , C=C , H−H , and C−H are 350,600,440 and 410kJ/mole respectively?

−130kJmol−1

−260kJmol−1

−400kJmol−1

−450kJmol−1