Question 16.CSGP.106: One kilomole of liquid oxygen, O2, at 93 K, and x kmol of ga......

One kilomole of liquid oxygen, O_2, at 93 K, and x kmol of gaseous hydrogen, H_2, at 25°C, are fed to a combustion chamber. x is greater than 2, such that there is excess hydrogen for the combustion process. There is a heat loss from the chamber of 1000 kJ per kmol of reactants. Products exit the chamber at chemical equilibrium at 3800 K, 400 kPa, and are assumed to include only H _2 O , H _2, and O.

a. Determine the equilibrium composition of the products and also x, the amount of H_2 entering the combustion chamber.
b. Should another substance(s) have been included in part (a) as being present in the products? Justify your answer.

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

x H _2+1 O _2 \rightarrow 2 H _2 O +( x -2) H _2

\text { (1) } 1 H _2 O \Leftrightarrow 1 H _2+1 O

shift -a +a +a and a > 0
Equil 2-a x-2+a a a < 2 and  n _{\text {tot }}= x + a

\text { (2) } 2 H _2 O \Leftrightarrow 2 H _2+1 O _2 \quad \ln K _2=-1.906

\text { (3) } 1 O _2 \Leftrightarrow 2 O \quad \ln K _3=-0.017

\ln K _1=0.5\left(\ln K _2+\ln K _3\right)=-0.9615 \Rightarrow K _1=0.3823

Equil.:  \frac{ K _1}{\left( P / P _{ o }\right)^1}=\frac{( x -2+ a ) a }{(2- a )( x + a )}=\frac{0.3823}{4}=0.95575

Energy Eq.:  Q + H _{ R }= H _{ P }, \quad Q =(1+ x )(-1000) \,kJ

Table A.9:  \Delta \overline{ h }_{ IG }^*=-5980 \,kJ / kmol [ or =0.922 \times 32(93-298.2)=-6054 \,kJ / kmol ]

Fig. D.2:  T _{ r }=93 / 154.6=0.601, \quad \Delta \overline{ h }_{ f }=-5.16 \times \overline{ R } \times 154.6=-6633

\begin{aligned}H _{ R }= & x (0+0)+1\left(0+\Delta \overline{ h }_{ IG }^*+\Delta \overline{ h }_{ f }\right)=1(-5980-6633)=-12613 \,kJ \\H _{ P }= & (2- a )(-241826+171981)+( x -2+ a )(0+119077) \\& + a (249170+73424)=119077 x +511516 a -377844 \\= & Q + H _{ R }=-1000-1000 x -12613\end{aligned}

Rearrange eq. to: x + 4.2599 a = 3.03331

Substitute it into the equilibrium eq.:  \frac{(1.03331+5.2599 a) a}{(2-a)(3.03331-3.2599 a)}=0.095575

Solve a = 0.198, LHS = 0.09547, x = 2.1898

y _{ H 2 O }=\frac{2- a }{ x + a }=0.755, \quad y _{ H 2}=\frac{ x -2+ a }{ x + a }=0.162, \quad y _{ O }=\frac{ a }{ x + a }=0.083

Other substances and reactions:  2 H _2 O \Leftrightarrow H _2+2 OH , \quad \ln K =-0.984

H _2 \Leftrightarrow 2 H , \quad: \ln K =0.201, \quad O _2 \Leftrightarrow 2 O ,: \quad \ln K =-0.017

All are significant as K’s are of order 1.

Related Answered Questions

Question: 16.CSGP.94

Verified Answer:

1 \,kmol \text { air }\left(0.78 N _2, 0.21...
Question: 16.CSGP.93

Verified Answer:

1 \,kmol \text { air }\left(0.78 \,N _2, 0....
Question: 16.CSGP.104

Verified Answer:

\begin{aligned}& \left( Z _1- Z _2\righ...
Question: 16.CSGP.101

Verified Answer:

Reaction:    2 CO +2 H _2 \Leftrightarrow C...