Displaystyle C-2-H-4-g-H-2-g-rightarrow C-2-H-6-g-Using the bond enthalpy data given below- calculate the enthalpy change the reaction-Bonddisplaystyle C-Cdisplaystyle C-Cdisplaystyle C-Hdisplaystyle H-HBond Enthalpy336-81 kJ - mol606-68 kJ - mol410-87 kJ - mol431-79 kJ - moldisplaystyle -120-08 kJ - moldisplaystyle -120-08 kJ - moldisplaystyle -240-16 kJ - moldisplaystyle -240-16 kJ - mol

Question

Toppr Admin · Accepted Answer

H-C2H4-H-C-C-4-xD7-H-C-x2212-H-606-68-4-xD7-410-87-2250-16kJ-mol

Bond	$C - C$	$C = C$	$C - H$	$H - H$
Bond Enthalpy	336.81 kJ / mol	606.68 kJ / mol	410.87 kJ / mol	431.79 kJ / mol

Bond	$C - C$	$C = C$	$C - H$	$H - H$
Bond Enthalpy	336.81 kJ / mol	606.68 kJ / mol	410.87 kJ / mol	431.79 kJ / mol

Bond	$C - C$	$C = C$	$C - H$	$H - H$
Bond Enthalpy	$336.81 k J / m o l$	$606.68 k J / m o l$	$410.87 k J / m o l$	$431.79 k J / m o l$

C2H4(g)+H2(g)→C2H6(g)Using the bond enthalpy data given below, calculate the enthalpy change for the reaction.BondC−CC=CC−HH−HBond Enthalpy336.81 kJ / mol606.68 kJ / mol410.87 kJ / mol431.79 kJ / mol−120.08 kJ / mol+120.08 kJ / mol−240.16 kJ / mol+240.16 kJ / mol

H(C2H4)=H(C=C)+4×H(C−H)=606.68+4×410.87=2250.16kJ/mol

$H (C_{2} H_{4}) = H (C = C) + 4 \times H (C - H)$

$= 606.68 + 4 \times 410.87 = 2250.16$ kJ/mol