THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<
THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<
9th Edition
ISBN: 9781266657610
Author: CENGEL
Publisher: MCG CUSTOM
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Chapter 16.6, Problem 93RP

A constant-volume tank contains a mixture of 1 mol of H2 and 0.5 mol of O2 at 25°C and 1 atm. The contents of the tank are ignited, and the final temperature and pressure in the tank are 2800 K and 5 atm, respectively. If the combustion gases consist of H2O, H2, and O2, determine (a) the equilibrium composition of the product gases and (b) the amount of heat transfer from the combustion chamber. Is it realistic to assume that no OH will be present in the equilibrium mixture?

a)

Expert Solution
Check Mark
To determine

The equilibrium composition of mixture of H2O, H2 and O2 at 5 atm pressure and 2,800 K.

Answer to Problem 93RP

The equilibrium composition of mixture of H2O, H2 and O2 at 5 atm pressure and 2,800 K is 0.944H2O+0.056H2+0.028O2.

Explanation of Solution

Write the stoichiometric reaction for dissociation of water.

H2+1/2O2H2O (I)

From Equation (I), infer that the stoichiometric coefficient for oxygen (vO2) is 0.5, and for both hydrogen (vH2) and water (vH2O) is 1.

Write the actual reaction for the combustion process.

H2+0.5O2xH2O+(1x)H2+(0.50.5x)O2 (II)

From Equation (II), infer that the equilibrium composition contains x amount of H2O (NH2O), (1x) amount of H2 (NH2), and (0.50.5x) amount of O2 (NO2).

Write the formula for total number of moles (Ntotal).

Ntotal=NH2O+NH2+NO2 (III)

Substitute x for NH2O, (1x) for NH2, and (0.50.5x) for NO2 in Equation (III).

Ntotal=x+(1x)+(0.50.5x)=1.50.5x

Write the equilibrium constant (kp) for the combustion process.

kp=NH2vH2NO2vO2NH2OvH2O(PNtotal)(vH2+vO2vH2O) (IV)

Here, pressure is P.

Conclusion:

Refer the table A-28, “Natural logarithm of equilibrium constants’, obtain the value of lnkp as 3.812 for the reported reaction at 2,800 K. Then the equilibrium constant for the reaction is 0.0221.

Substitute 0.0221 for kp, x for NH2O, (1x) for NH2, (0.50.5x) for NO2, 1.50.5x for Ntotal,5 atm for P, 1 for both vH2, and vH2O, and 0.5 for vO2 in Equation (IV).

0.0221=(1x)(0.50.5x)0.5x(5atm1.50.5x)(1+0.51)0.0221x=(1x)(0.50.5x)0.5(5atm1.50.5x)(0.5)0.000488x2=(1+x22x)(0.50.5x)(5atm3(0.50.5x))

4.998x210x+5=0x=0.944

Substitute 0.944 for x in Equation (II).

H2+0.5O2xH2O+(1x)H2+(0.50.5x)O2H2+0.5O20.944H2O+(10.944)H2+(0.50.50(.944))O2H2+0.5O20.944H2O+0.056H2+0.028O2

Thus, the equilibrium composition of mixture of H2O, H2 and O2 at 5 atm pressure and 2,800 K is 0.944H2O+0.056H2+0.028O2.

b)

Expert Solution
Check Mark
To determine

The amount of heat released per kg of hydrogen.

Answer to Problem 93RP

The amount of heat released per kg of hydrogen is 132,600kJ/kmolH2.

Explanation of Solution

Write the energy balance equation for the combustion process.

Qout=NP(h¯f+h¯h¯RuT)PNR(h¯fRuT)R={NH2O(h¯f+h¯h¯RuT)H2O+NH2(h¯f+h¯h¯RuT)H2+NO2(h¯f+h¯h¯RuT)O2NH2(h¯fRuT)H2NO2(h¯fRuT)O2} (V)

Here, heat released during combustion is Qout, number of moles of product is NP, number of moles of reactants is NR, number of moles of H2O is NH2O, number of moles of H2 is NH2, number of moles of O2 is NO2, enthalpy at reference state is h¯f, sensible enthalpy at specified state is h¯, sensible enthalpy at reference state is h¯, universal gas constant is Ru, and temperature is T.

Conclusion:

Refer the table A-26, “Enthalpy of formation table”, obtain the enthalpy of H2O (h¯f)H2O, O2 (h¯f)O2,and H2 (h¯f)H2 as 241,820kJ/kmol, 0, and 0.

Refer the table A-18, “Ideal gas properties of hydrogen gas”, obtain the following properties of hydrogen gas at different temperature.

Enthalpy of hydrogen gas at 2800 K, (h¯)H2=89,838kJ/kmol.

Enthalpy of hydrogen gas at 298 K, (h¯)H2=8,468kJ/kmol.

Refer the table A-19, “Ideal gas properties of oxygen gas”, obtain the following properties of oxygen gas at different temperature.

Enthalpy of oxygen gas at 2800 K, (h¯)O2=98,826kJ/kmol.

Enthalpy of oxygen gas at 298 K, (h¯)O2=8,682kJ/kmol.

Refer the table A-19, “Ideal gas properties of water vapor”, obtain the following properties of water vapor at different temperature

Enthalpy of water vapor at 2800 K, (h¯)H2O=125,198kJ/kmol.

Enthalpy of water vapor at 298 K, (h¯)H2O=9,904kJ/kmol.

Substitute 0 for (h¯f)H2, 89,838kJ/kmol for (h¯)H2, 8,468kJ/kmol for (h¯)H2, 0 for (h¯f)O2, 98,826kJ/kmol for (h¯)O2, 8,682kJ/kmol for (h¯)O2, 241,820kJ/kmol for (h¯f)H2O, 125,198kJ/kmol for (h¯)H2O, 9,904kJ/kmol for (h¯)H2O 0.944 for NH2O, 0.056 for NH2, 0.028 for NO2, 8.314kJ/kmolK for Ru, and 2800 K for T in Equation (V).

Qout={0.944(241,820kJ/kmol+125,198kJ/kmol9,904kJ/kmol(8.314kJ/kmolK)(2800K))+0.056(0+89,838kJ/kmol8,468kJ/kmol(8.314kJ/kmolK)(2800K))+0.028(0+98,826kJ/kmol8,682kJ/kmol(8.314kJ/kmolK)(2800K))1(0(8.314kJ/kmolK)(298K))0.5(0(8.314kJ/kmolK)(298K))}=132,600kJ/kmolH2

Qout=132,600kJ/kmolH2

Thus, the amount of heat released per kg of hydrogen is 132,600kJ/kmolH2.

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