The value of enthalpy change `(DeltaH)` for the reaction
`C_(2)H_(5)OH (l)+3O_(2) (g) rarr 2CO_(2) (g) +3H_(2)O (l)`
at `27^(@)C` is `-1366.5 kJ mol^(-1)`.
The value of internal energy change for the above reactio at this temperature will be

Options

A. `-1371 . 5 K J`
B. `-1369.0 K J `
C. `-1364.0 K J`
D. `-1361 . 5 K J

Answer

C

Explanation

Relation between `Delta H` (enthalpy change) and `Delta E` (internal energy change) is
`Delta H = DeltaE + Delta_(g) nRT`
Where, `Deltan_(g)` = (moles of gaseous products) – (moles of gaseous reactants)
For the given reaction,
`Delta n_(g) = 2- 3 = -1`
`implies -1366.5 = Delta E - 1 xx 8.314 xx 10^(-3) xx 300`
`:. " "Delta E = -1364.0 kJ mol^(-1)`.