Chemical BondingEasy
Question
If at 298 K the bond energies of C-H, C-C, C = C and H-H bonds are respectively 414, 347, 615 and 435 kJ mol-1, the value of enthalpy change for the reaction H2C = CH2(g) + H2(g) → H3C - CH3(g) at 298 K will be
Options
A.- 250 kJ
B.+ 125 kJ
C.- 125 kJ
D.+ 250 kJ
Solution
CH2 = CH2 (g) + H2(g) → CH3 - CH3
ᐃH = 1(C = C) + 4(C - H) + 1(H-H) -1(C - C) - 6(C - H) = 1(C = C) + 1(H - H) - 1(C - C) - 2(C - H)
= 615 + 435 - 347 - 2 × 414 = 1050 - 1175 = - 125kJ.
ᐃH = 1(C = C) + 4(C - H) + 1(H-H) -1(C - C) - 6(C - H) = 1(C = C) + 1(H - H) - 1(C - C) - 2(C - H)
= 615 + 435 - 347 - 2 × 414 = 1050 - 1175 = - 125kJ.
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