From the following bond energies: H - H bond energy: 431.37 kJ mol -1 C = C bond energy: 606.10 kJ mol -1 C - C bond energy: 336.49 kJ mol -1 C - H bond energy: 410.50 kJ mol -1 Enthalpy for the reaction, will be

Question

From the following bond energies: H - H bond energy: 431.37 kJ mol -1 C = C bond energy: 606.10 kJ mol -1 C - C bond energy: 336.49 kJ mol -1 C - H bond energy: 410.50 kJ mol -1 Enthalpy for the reaction, will be (A) 553.0 kJ mol-1 (B) 1523.6 kJ mol-1 (C) -2 43.6 kJ mol-1 (D) -120.0 kJ mol-1

Tardigrade Admin · Accepted Answer

Δ H = dissociation energy of reactant - Bond dissociation of energy of product. Δ H =(606.10+4 × 410.5+431.37)-(6 × 410.50+336.49) =-120.0 kJ / mol

Q. From the following bond energies :
$H - H$ bond energy $: 431.37 k J m o l^{- 1}$
$C = C$ bond energy $: 606.10 k J m o l^{- 1}$
$C - C$ bond energy $: 336.49 k J m o l^{- 1}$
$C - H$ bond energy $: 410.50 k J m o l^{- 1}$
Enthalpy for the reaction, will be

$553.0 k J m o l^{- 1}$

$1523.6 k J m o l^{- 1}$

$- 243.6 k J m o l^{- 1}$

$- 120.0 k J m o l^{- 1}$

$Δ H =$ dissociation energy of reactant $-$ Bond dissociation of energy of product.

$Δ H = (606.10 + 4 \times 410.5 + 431.37) - (6 \times 410.50 + 336.49)$

$= - 120.0 k J / m o l$

Q. From the following bond energies : H−H bond energy :431.37kJmol−1 C=C bond energy :606.10kJmol−1 C−C bond energy :336.49kJmol−1 C−H bond energy :410.50kJmol−1 Enthalpy for the reaction, will be

553.0kJmol−1

1523.6kJmol−1

−243.6kJmol−1

−120.0kJmol−1