1 mole of H2 and 0.5 mole of O

1 mole of H2 and 0.5 mole of O2 gases at 298K are introducedinto an insulating reaction chamber, which has a volume of 10liters and already contains 2 moles of water. Reaction between H2and O2 is triggered by a spark. Answer the following questions: i)Is the reaction product in the chamber water or steam? What is itstemperature? ii) The reaction product pushes a piston through areversible adiabatic expansion until the pressure in the reactionchamber drops to 1 bar. How much work does the system do? Usefuldata: enthalpy of reaction of H2O ΔfH(H2O)(298K) = -240 kJ/mol,heat capacity of liquid water C(v,l) = 75 J/(mol·K), heat capacityof steam C(v,g) = 28 J/(mol·K), enthalpy of vaporization of H2OΔvapH(H2O) = 41 kJ/mol. Assume that liquid water boils at 373K andsteam behaves like an ideal gas in the considered situation.


The reaction is H2+0.5O2—> H2O

So due to reaction of 1 mole of H2 and 0.5 mole of O2, 1 mole ofwater is formed and 2 moles of water is already there in thereactor. Hence total moles of water =1+2= 3moles

depending upon the enthalpy of reaction one can say whether theproduct is liquid or vapor.

Enthalpy of reaction =-240 Kj/mol since only one mole of productis produced, heat generated during the course of reaction= 240 Kj=240*1000 joules= 2,40,000 joules

This heat has to be taken by 3 mole of water to rise itstemperature

Sensible heat for rising the temperature of water from 298 k to373.15 ( which is the boiling point of water)

= number of moles* specific heat of water* temperaturedifference= 3*75*(373.15-298)=16908.75 joules

Latent heat= 3* 41*1000 joule/mole= 123000

Total heat required for converting 3 moles of liqudi water at298K to water vapr at 393.15K= 16908.75+123000=1,39,909 joules

This total heat is less than 2,40,000 joules. Hence all thewater will be in the form of steam only.

Additional heat = 2,40,000-139909=100091.3 joules

This heat will be used to rise the temperature of water vaprfrom 100 deg.c =373.15 K to T (unknwon)


T-273.15= 1191.6


b) T=1464.75 K and n= number of moles = 3 moles R=0.08206L.atm/mole.K

from gas law P= nRT/V= 3*0.08206*1464.75/10 =36.05 atm

for adiabatic expansion delU =Q+W

Q= 0 and delU= W

from P1V1Y= P2V2Y, the final volume at1bar need to be claculated

36.05*101.33 = 0.9869* V2 1.33

213.8 =0.9869* V2 1.33

V2= 56.5 L

Work done = (P2V2- P1V1)/ (Y-1)= (1*56.5-36*10)/(1.33-1)=-920L.atm =-920*101.3 j/mole=-93196 joules


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