Question

Calculate the standard internal energy change for the formation of water at 298 K. The (5 ) standard enthalpy of formation of H2O (l) at 298 K is  285.8 kJ mol

Answer 1

H2 (g) + 1/2 * O2 (g) ==> H2O (l)   at 298°C

Initially H2 & O2 are gases. Let us assume that the conditions are STP.  Then after the reaction the volumes comes down to 18 ml of  water for 1 mole. Initially 1 mole of H2 has 22.4 liters and half mole of Oxygen occupies 11.2 liters.   Final pressure changes as volume changes and temperature remains constant.

Work done
$W=-\int {P\ dV}=-RT \int {\frac{1}{V}} \, dV\\\\=-RT\ Ln \frac{V2}{V1}\\\\=-8.314*298*\ Ln\frac{18}{33600}\ J\\\\=18660.85 J$ 

We know from the law of thermodynamics:
ΔH = ΔU + W
-285.8 kJ / mol  =  ΔU + 18.66085 kJ 

=> ΔU = -304.46 kJ