The enthalpy changes for the following reactions at `298k` and `1atm` are given below:
a. `CH_(3)COOH(l)+2O_(2)(g) rarr 2CO_(2)(g) +2H_(2)O(l),DeltaH =- 874 kJ`
b. `CH_(3)CH_(2)OH(l)+3O_(2)(g)rarr 2CO_(2)(g)+3H_(2)O(l),DeltaH =- 1363 kJ`
Calculate the internal enegry changes for these recaitons.

Answer

(i)`-874` kJ
`-1360.5`kJ

Explanation

`DeltaH = DeltaU + Deltan_(g)RT`
(i)`CH_(3)COOH(l) + 2O_(2)(g) to 2CO_(2)(g) + 2H_(2)O(l),`
`DeltaH = -874 kJ`
`Deltan_(g) = 2 - 2 = 0`
`therefore - 874 kJ = DeltaU + 0 xx RT or DeltaU = -874 kJ`
(ii)`CH_(3)CH_(2)OH(l) + 3O_(2)(g) to 2CO_(2)(g) + 3H_(2)O(l), `
`DeltaH = -1363kJ`
`Deltan_(g) = 2 - 3 = -1`
`-1363 kJ = DeltaU - 1 xx 8.314 xx 10^(-3) xx 298`
`DeltaU = -1363 + 2.47 = -1360.53 kJ`.