Question

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

Answer 1

The reaction is H2+0.5O2---> H2O

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

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

Enthalpy of reaction =-240 Kj/mol since only one mole of product is 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 its temperature

Sensible heat for rising the temperature of water from 298 k to 373.15 ( which is the boiling point of water)

= number of moles* specific heat of water* temperature difference= 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 at 298K 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 the water 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 vapr from 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=0.08206 L.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 P1V1^Y= P2V2^Y, the final volume at 1bar need to be claculated

36.05*10^1.33 = 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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