# 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.

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

100091.3=3*28*(T-373.15)

T-273.15= 1191.6

T=1191.6+273.15=1464.75K

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

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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