a. Cis-`2`-butene `rarr` trans-`2`-butane, `DeltaH_(1)`
b. Cis-`2`-butane `rarr1`-butene, `DeltaH_(2)`
c. Trans-`2`-butene is more stable than `C` is-`2`-butene
d. Ethalpy of combustion of `1`-butene, `DeltaH =- 649.8 kcal//mol`
e. `9DeltaH_(1) +5DeltaH_(2) = 0`
f. Enthalpy of combustion of trans `-2`-butene, `DeltaH =- 647.1 kcal mol^(-1)`. Calculate `DeltaH_(1)`and `DeltaH_(2)`?

Explanation

a. `cis-2`-Butene `rarrtrans-"Butene", DeltaH_(1)`
b. `cis-2-"Butene" rarr 1-"Butene", DeltaH_(2)`
c. `trans-2-`Butane
`+6O_(2)(g) rarr 4CO_(2)(g) +4H_(2)O(l), DeltaH =- 647.1 kcal mol^(-1)`
d. `1-"Butene"`
`+6O_(2)(g) rarr 4CO_(2)(g) +4H_(2)O(l). DeltaH =- 649.8 kcal mol^(-1)`
Subtracting equation (iv) from (iii)
`-647.1 +649.8 = Delta_(f)H^(Theta) (1-"Butene") -Delta_(f)H(trans-2-"butene")`
`2.7 = Delta_(f)H (1-"Butene") -Delta_(f)H (trans-2-"butene")`
Subtracting equation (i) from (ii), we get
`DeltaH_(2) - DeltaH_(1) = Delta_(f)H(1-"Butene") -Delta_(F)H(trans-2-"butene") = 2.7`
Now, `9DeltaH_(1) +5DeltaH_(2) = 0`
`DeltaH_(2) - DeltaH_(1) = 2.7`
On solving, we get
`DeltaH_(1) =- 0.96 kcal`
`DeltaH_(2) = 1.74 kcal`