Question

For the chemical reaction C2H5OH(ℓ) + 3O2 (g) → 2CO2 + 3H2O (ℓ), if change in internal energy at 298K is 1368 KJ/mole, the change in enthalpy will be

Answer 1

-1373KJ mole, - 1,T=298k is my answer

Answer 2

Answer: $\Delta H=-1109.6kJ/mol$

Explanation:-

$C_2H_5OH(l)+3O_2 (g)\rightarrow 2CO_2(g)+3H_2O(l)$

$\Delta H=\Delta U+\Delta n_gRT$

$\Delta H$= enthalpy change = ?

$\Delta U$= internal energy change = 1368 kJ/mol

$\Delta n_g$= number of moles of gaseous products -  number of moles of gaseous reactants = (2-3)= -1

R= gas constant = 8.314 J/Kmol

T = temperature = 298 K

$\Delta H=1368+(-1)\times 8.314\times 298$

$\Delta H=-1109.6kJ/mol$

The change in enthalpy will be -1109.6 kJ.