Question

Calculate the enthalpy of the following reaction. CH2=CH2+H2—> CH3–CH3. The bond energies of C-H,C-C,C=C and H-H are 414,347,615 and 435 KJ/mol respectively .

Answer 1

Answer:

Calculate the enthalpy of the following reaction. CH2=CH2+H2—> CH3–CH3. The bond energies of C-H,C-C,C=C and H-H are 414,347,615 and 435 KJ/mol respectively .

Answer 2

tAnswer:

given =  CH =CH₂ + H₂ → CH₃---CH₃

BOND ENERGIES OF,

C-H = 414 KJ/Mol

C-C = 347 KJ/mol

C=C = 615 KJ/mol

H-H = 435 KJ/ mol

to find =  enthalpy of given reaction.

solution :-  

ΔH =  ΔH ( reactant ) -  ΔH (Product)

ΔH = 4× ( C-H )  +1×  C=C -  1×( H-H ) -6 C-H +C-C

ΔH =4×414 +615 +435  - 6× 414 +  349

ΔH =  -125

the enthaply of the given reaction is -125.