For complete combustion of ethanol, `C_(2)H_(5)OH(l) + 3O_(2)(g) rarr 2CO_(2)(g) + 3H_(2)O (l)`, the amount of heat produced as measured in bomb calorimeter , is `1364.47 kJ mol^(-1)` at `25^(0)C`. Assuming ideality the Enthalpy of combustion, `Delta_(C)H`, for the reaction will be: `(R = 8.314 kJ mol^(-)`).

Options

A. `-1 3 6 . 9 5 K J mol^(-1)`
B. `-1 3 6 1 .9 5 K J mol^(-1)`
C. `1 4 6 0 . 5 0 K J mol^(-1)`
D. `-1 3 5 0 . 5 0 K J mol^(-1)`

Answer

A

Explanation

`C_(2)H_(5)H(l) + 3O_(2) (g) to 2CO_(2)(g) + 3H_(2)O(l)`
Amount of heat produced in bomb calorimeter.
`Delta U = -1364.47 kJ mol^(-1)`
Enthalpy of a combustion reaction is
`DeltaH = DeltaU + Deltan_(g) RT`
Where, `DeltaU` = internal energy
`Delta n_(g)` = moles of gas (products-reactants)
`Delta R` = Gas constant
T = Temperature in K
As per equation,
`Delta n_(g) = 2 - 3 = -1`
`T = 25^@C = 25 + 273 K`
`implies T = 293 K`
Thus, `Delta H = -1364.47 + [((-1) xx 8.314 xx 298)/(1000)]`
`Delta H = -1364.47 - 2.477 = -1366.947 kJ mol^(-1)`
Hence, the enthalpy of combustion, `Delta_(C) H` for the given reaction will be `-1366.947 kJ mol^(-1)`.