Answer:
(a) -134 kJ (b) -136 kJ
Explanation:
a) Using the accounting of the enthalpy of broken and formed bonds we will have the answer for this question ( remember the enthalpy of bonds broken are positive and those formed are negative)
ΔHrxn(kJ) = 6H (C-H) + 1 (C-C)+ 1H (H) + 1(C=C) + 4(C-H)
= 6(-411) +(-346) +432 + 602 + 4(411)
= - 134 kJ
(b) ΔHrxn = ΔHºf(C2H6) - ΔHºf (C2H4) = -83.75 kJ- 52.3 kJ = -136 kJ
( ΔHºf (H2)) = 0)
