Calculate the enthalpy of formation of `Delta_(f)H` for `C_(2)H_(5)OH` from tabulated data and its heat of combustion as represented by the following equaitons:
i. `H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(g), DeltaH^(Theta) =- 241.8 kJ mol^(-1)`
ii. `C(s) +O_(2)(g) rarr CO_(2)(g),DeltaH^(Theta) =- 393.5kJ mol^(-1)`
iii. `C_(2)H_(5)OH (l) +3O_(2)(g) rarr 3H_(2)O(g) + 2CO_(2)(g), DeltaH^(Theta) =- 1234.7kJ mol^(-1)`
a. `-2747.1 kJ mol^(-1)` b. `-277.7 kJ mol^(-1)`
c. `277.7 kJ mol^(-1)` d. `2747.1 kJ mol^(-1)`

Options

A. `-2747.1kJ" "mol^(-1)`
B. `-277.7kJ" "mol^(-1)`
C. `277.7kJ" "mol^(-1)`
D. `2747.1kJ" "mol^(-1)`

Answer

B

Explanation

Required equation:
`2C(s)+3H_(2)(g)+1//2O_(2)(g)toC_(2)H_(5)OH(l)`
`2C(s)+2O_(2)(g)to2CO_(2)(g),DeltaH^(@)=-2xx393.5kJ`
`3H_(2)(g)+(3)/(2)O_(2)(g)to3H_(2)O(g),DeltaH^(@)=3xx241.8kJ`
`underline(3H_(2)O(l)+2CO_(2)(g)toC_(2)H_(5)OH(l)+3O_(2)(g)," "DeltaH=1234.7kJ)`
On adding `underline(2C(s)+3H_(2)(g)+(1)/(2)O_(2)(g)toC_(2)H_(5)OH(l)," "DeltaH^(@)=-277.7kJ" "mol^(-1))`