One mole of h2o(g) at 1.00 atm and 100.°c occupies a volume of 30.6 l. when one mole of h2o(g) is condensed to one mole of h2o(l) at 1.00 atm and 100.°c, 40.66 kj of heat is released. if the density of h2o(l) at this temperature and pressure is 0.996 g/cm3, calculate δe for the condensation of one mole of water at 1.00 atm and 100.°c.

ΔE (of water)= -37,56 KJ

Explanation:

Using the first law of thermodynamics:

ΔE= Q - W = ΔH - P(Vf - Vi)

- The heat released by water in its condensation diminishes its internal energy --> Negative contribution

- The work done to compress the water should increase its internal energy --> positive contribution (Vf< Vi) , Vf= mf/Df

W= P(Vf - Vi) = 1 atm * (101325 Pa/atm) * [  1/(0,996gr/cm3)* 18 gr/mole*1 mole - 30,6 L * 1m3/1000L]] * 1 KJ/1000J = -3,01 KJ

ΔE= Q - W = (-40,66 KJ/mole)*1 mole - (-3,01 KJ) = -40,66 KJ + 3,01 KJ = (-37,56 KJ)

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :