Question

COmbusting 44.8L of H2(g) with 22.4L of O2(g) (both at STP) results in the formation of 44.8L of H2O(g) if the reaction is maintained at the same condition (STP) and goes to completion. Given the heat of formation of water (delta Hf=-241.8 kJ/mol) calculate: a) the PV work b) the heat evolved c) the change in internal energy of the system and d) the change in internal energy of the surroundings.

Answer 1

2H₂ + O₂ ----------> 2H₂O

Given, conversion of Liters to moles

moles of H2 =

Moles of O2 =

Moles of H2O =

Here moles are according balanced equation

So enthalpy of reaction =

ENthalpy of reaction = -241.8 kJ

Work done by the system

Pressure = 1 atm

Change in volume =

= -1(-22.4) L.atm

W= 22.4 L.atm

W = 22.4(101.3) =2269.1 J = 2.27 kJ

Heat evolved = enthalpy of reaction = 241.8 kJ

But at constant pressure ,

q = 241.8 kJ

Change in the internal energy of the system =

Change internal energy of the syetem = 239.5 kJ

Change internal energy of the surroundings =