Chemistry Problem_ Thermochemistry?

Question

The standard enthalpy of formation of H20 at 298 K is -285.8 kJ/mol. Calculate the change in internal energy for the following reaction at 298 K and 1.00 atm:

2H20(l)---) 2H2(g) +O2(g) ^E =?

hint:use the ideal gas law to derive an expression in terms of n, R, and t

Answer 1

The standard enthalpy of the reaction 2 H2O->2 H2 + O2 is  &Delta Hr= 0+0 - 2* -285.8 KJ/mol= 571.16 KJ/mol.

The internal energy &Ur can be derived from

d Hr= d Ur+d (p*V)

since p*V= nRT we get d(p*V)=nR dT

-> dHr = dUr +nR dT

or &Delta Hr = &Delta Ur +nR &Delta T

since &Delta T = 0 // T=298 K

we find &Delta Hr = &Delta Ur