Gas, Liquid and Solid Mixtures

5. Gas, Liquid and Solid Mixtures

Consider the following chemical reactions

CaCO 3 (s) → CaO(s) + CO 2 (g),

H 2 O(l) + CO(g) → CO 2 (g) + H 2 (g), and

(52) All three reactions consider species in two separate phases called heterogeneous reactions. For

CaO(s) + SO 2 (g) + 1/2 O 2 (g) → CaSO 4 (s).

reaction given by Eq. (52) the equilibrium relation follows from Eq. (35a), i.e.,

0 f CaSO 4 () s (,)ˆ TP α CaSO 4 () s / f CaSO 4 () s (,) T KT 1 () =

f SO 2 () g (,)ˆ TP α SO 2 () g f O 2 () g (,)ˆ TP α O 2 () g f CaO s () (,)ˆ TP α CaO s ()

f SO 2 () g (,) T 1 f O 2 () g (,) T 1 f CaO s () ( T, ) T 1 If CaSO 4 and CaO are fully mixed at the molecular level, assuming that the solid mixture is an

ideal solution, α ˆ CaSO 4 () s =X CaSO 4 () s , and α ˆ CaO(s) =X CaO(s) , where X CaSO 4 () s +X CaO(s) =1.

(54) For the ideal solution f k(s) (T, P) ≈f k(s) (T, 1). Assuming ideal gas behavior for SO 2 ,

α ˆ SO 2 =X SO 2 , α ˆ O 2 =X O 2 ,f SO 2 (T, P) = P, and f O 2 (T, P) = P. (55) Therefore, Eq. (53) assumes the form

K –3/2 (T) = (X

CaSO 4 () s /(X SO 2 X O 2 X CaO(s) ) (P/1) .

i. Example 9 Determine the relations between the partial pressures and temperature for the follow- ing scenarios: pure H 2 SO 4 (l) dissociating upon evaporation, H 2 SO 4 (l)

→H 2 O(g) + SO 3 (g), and an ideal mixture of 40% volatile H 2 SO 4 (l) and 60% nonvolatile liquid or

solid participating in the same reaction. Assume that

g = –690013 kJ kmole 2 –1 4 H SO ,

g –1 HOg 2 () = –228572 kJ kmole , g SO g 3 () = –371060 kJ kmole at 298 K and 1 bar. Determine (p HOg 2 () )(p SO g 3 () ) at 298 K for pure H 2 SO 4 (l) and for an ideal mixture of

40% volatile H 2 SO 4 (l) and 60% nonvolatile liquid.

Solution Pure H 2 SO 4 (l) The problem involves a mixture of phases. We will select the standard state to be the

liquid state for H 2 SO 4 (l) . Then from Eq. (35a),

K ν (T) = Π(f

k (T,P) α ˆ k /f k (T, 1 bar)) k .

The gaseous species are assumed to be ideal so that

f k(g) (T, P) = P, f k(g) (T, 1 bar) = 1, and α ˆ k(g) (T, P, X k )=X k . In the liquid phase at 1 bar

f 2 4 H SO (T, P)/f 2 4 H SO (T, 1 bar) ≈ 1.

Further, since the liquid is pure, α ˆ

2 4 H SO = 1, and ˆ K o (T) =(p

HOg 2 () /1) 1 (p SO g 3 () /1) .

At 298 K, when X 2 4 H SO =1

(p

)(p

) = 1.44 ×10 HOg –16 2 () SO g 3 () ,

which indicates that the partial pressures are very low, i.e., there is negligible disso- ciation. Ideal Liquid Mixture:

Since the liquid phase is in an ideal mixture, the activity of H 2 SO 4 (l) = X 2 4 H SO , and K o (T) = (P X

/1) HOg 1 2 () (P X SO g 3 () /1) 1 /X 2 4 H SO

HOg = (p 2 () /1)(p SO g 3 () /1)/X 2 4 H SO , i.e.,

(p

HOg 2 () )(p SO g 3 () )=X 2 4 H SO K (T).

The Gibbs energy change

∆G o (T) = g HOg 2 () + g SO g 3 () – g 2 4 H SO

= –228572 –371060 + 690013 = 90381 kJ kmole –1 , and ln K o (T) = – ∆G o (T)/ ( R T) = –36.48, i.e.,

K o (T) = 1.44 ×10 –16 . At 298 K, when X = 0.4,

SO g 3 () ) = 0.4 ×1.44×10 = 0.576 ×10 . Remarks During the vaporization of H 2 SO 4 (l) it is possible to produce H 2 O, SO 2 , SO 3 , and O 2 ,

rather than H 2 SO 4 (g). The pertinent reactions are

H 2 SO 4 (l) →H 2 O(g) + SO 2 (g) + 1/2 O 2 (g), and

H 2 SO 4 (l) →H 2 O(g) + SO 3 (g)

At equilibrium, can you determine the SO 2 and SO 3 concentrations? j. Example 10

Consider the water gas shift reaction H 2 O(l) + CO →H 2 (g) + CO 2 (g). Determine the equilibrium constant for the reaction at 298 K, treating the gaseous species as ideal. Solution Since CO, H 2 and CO 2 are treated as ideal gases, f k = P, ˆa H2O = 1, f H2O(l) (T,P) ≈f H2O(l) (T,1) and for others α ˆ k =X k so that Eq. (35a) transforms to

K o (T) = (p H 2 /1)(p CO 3 /1)/((p CO /1) X HO 2 ) = (p H 2 /1)(p CO 3 /1)/(p CO /1)

(A) Using tabulated values,

= exp (–( o g + g – g o

H 2 CO 2 CO – g HO 2 )/ R T).

HO 2 (l) = h HO 2 –T s HO 2

×69.95 = –306675 kJ kmole = –285830 – 298 –1 of H 2 O(l) ,

g o CO = –110530 – 298 ×197.56 = –169403 kJ kmole –1 of CO,

g o = –393520 – 298 ×213.7 = –457203 kJ kmole CO –1 2 of CO 2 , and

g o 2 = 0 – 298 ×130.57 = –38910 kJ kmole H –1 of H 2 .

Therefore, ∆G o = –38910 – 457203 – (–169403 – 306675)

= –20035 kJ kmole –1 of CO, and K o (298 K) = exp (20,035 ÷(8.314×298)) = 3250.4.

Remarks Since K o (298 K) is extremely large, X CO 3 X H 2 /X CO is also large, and, consequently,

the value of X CO at chemical equilibrium is extremely small. Therefore, if CO gas is bubbled through a vast reservoir of H 2 O(l) at 298 K, very little unreacted CO is left over. Note that the results pertain only to an equilibrium condition. However, the time

scale required to reach it may be inordinately large. k. Example 11

Consider the reaction of SO 3 (g) with CaO(s), a process that is used to capture the SO 3 released during the combustion of coal, i.e., CaO(s) + SO 3 (g) → CaSO 4 (s). Determine the equilibrium relation assuming that the sulfates and CaO are mixed at the molecu- lar level (i.e., they are mutually soluble) and are unmixed. What is the partial pressure

of SO at 1200 K if K o (1200 K) = 2.93 ×10 7 3 .

Solution Assume the standard for solids to be solid and for gases to be ideal gases at P = 1 bar and use the approximation that f (s)(T,P) k ≈f (s)(T, 1bar). k

Since gas SO 3 behaves like an ideal gas,

α ˆ SO 3 =X SO 3 ,f SO 3 (T, P) = P.

The solid phase contains both CaSO 4 and CaO. Assuming the ideal solution model for the solid phase,

α ˆ CaSO 4 =X CaSO 4 , and α ˆ CaO =X CaO =1–X CaSO 4 .

The equilibrium relation for the reaction is,

K o (T) = X CaSO 4 /(X CaO (P X SO 3 )/1).

If the solids are not mixed at the molecular level, they exist separately. Therefore,

X CaSO 4 =X CaO = 1, and K o (T) = 1/((P X

SO 3 )/1).

In the unmixed case,

2.93 ×10 7 = 1/(p

SO 3 /1), i.e., p SO 3 = 0.41 ×10 bar.

Remarks If the pressure P = 1 bar, then for the unmixed case X SO 3 = p SO /P = 0.41 ×10 –8 3 or

0.0041 ppm (parts per million). If the pressure is isothermally increased, the value of p SO 3 remains unchanged, but X SO 3 decreases, i.e., more of the sulfate will be formed.

In many instances, in power plants SO 2 released due to coal combustion is allowed to

react with lime in order to produce sulfates according to the following reaction CaO(s) + SO 2 + 1/2O 2 → CaSO 4 (s).

The equilibrium relation for this reaction is K o (T) = 1/((p SO /1)(p /1) 1/2 2 O 2 )) = (1/(X SO 2 X O 1/2 2 )) (P/1) –1/2 .

Increasing the pressure at constant T, causes X SO 2 to decrease so that a lesser amount of SO 2 will be emitted, i.e., more SO 2 is captured from the combustion gases. l. Example 12 One kmole each of C(s) and O 2 enter a reactor at 298 K. The species CO, CO 2 , and O 2

leave the reactor at 3000 K and 1 bar at equilibrium. Find the value of the equilibrium composition at the exit. What is the heat transfer across the boundary? What will hap- pen if the inlet stream is altered to contain 1/2 kmole of oxygen and one kmole of

CO? Also explain what happens if the outlet contains C(s), CO, and O 2 .

Solution The overall chemical reaction is

C(s) + O 2 → a CO 2 + b CO + c O 2 (A) The species leaving the reactor are in an equilibrium state so that the following reac-

tion must be in equilibrium, namely,

(B) From an atom balance,

CO 2 → CO + 1/2 O 2 .

C atoms 1 = a + b (C) O atoms: 2 = 2a + b + 2c.

(D) Therefore,

b = 1 – a, and c = (1 – a)/2. (E,F) The total moles leaving the reactor are N = a + b+ c = (3 – a)/2.

(G) The exit equilibrium condition requires that K o (T) = p CO p O2 _ /p CO2

(H) For the carbon dioxide dissociation reaction at 3000 K,

K o (3000 K) = 0.327. Since,

X CO = b/N, X CO 2 = a/N, X O 2 = c/N, and p k =X k ×1 bar, Solving the three unknowns a,b and c from Eqs. (B), (C) and (H)

a = 0.563, b = 0.437, and c = 0.219. Applying the First Law

(I) Under steady state, dE cv /dt =0, and there is no work transfer. Thus, for every kmole of

dE cv /dt = ˙ Q cv – W ˙ cv + Σ ,k N ˙ ik h i,k – Σ k N ˙ e,k h e,k ,

C(s), Q ˙

q = ˙ == 0.563 h CO 2 (3000 K) + 0. 437 h CO (3000 K) + 0. 219 h O N

2 (3000 K) –

cs ()

h C(s) (298 K) – 1/2 h O 2 (298 K) = 121426 kJ per kmole of C(s) consumed. (J) If the inlet stream is altered to contain 1/2 kmole of oxygen and one kmole of CO, the

atom balance remains unchanged. Therefore, the outlet composition will remain un- altered. However, the heat transfer will change, since the inlet stream containing CO

and O 2 has a lower enthalpy as compared to the mixture of C(s) and O 2 . Hence, the value of Q will be lower. If CO,O2 and C(s) are present at the outlet, the overall chemical reaction is

(K) There are two atom balance equations, and we will also consider the following reac-

C(s) + 1 O 2 → b CO + c O 2 + d C(s)

tion to be in equilibrium, i.e., C(s) + 1/2 O 2 → CO, so that (L)

K o (T) = p CO /1/((p O 2 /1) 1/2 (f C(s) (T,P)/f C(s) (T,1)))

Since (f C(s) (T,P) ≈f C(s) (T,1)),

(M) Hence, from O atom and C atom balances

K o (T) = p CO /1/((p O 2 /1) 1/2 ) = (X CO /X O 2 ) (P/1) 1/2 .

2= b+2c, b + d = 1; thus c = (2-b)/2, d = (1-b), and N = b+c = (b + 2)/2, so that

(b = 2/{(P/2) 1/2 /K o (T) + 1})< 1, and d = 1 – b.

For the C(s) + 1/2 O 2 → CO,

K 6.4 (3000 K) = 10 . Since K o (3000 K) is relatively large, Eq. (M) suggest that X O 2 ≈ 0, i.e., almost all of

the oxygen is consumed and converted into product. m. Example 13

Let the number of kmole of O 2 entering a reactor equal the value a, while the number of moles of O 2 , CO, CO 2 , and C(s) leaving that reactor equal N O 2 , N CO , N CO 2 , and

N C(s) . Assume the following reactions (respectively, A and B) to be in equilibrium: C(s)+ 1/2 O 2 → CO, and CO 2 → CO + 1/2 O 2 . Determine N CO (T) and minimum amount of carbon(s) that should enter the reactor so as to maintain equilibrium at the

exit. Solution

(C) K

O atom balance: 2a = 2 N O 2 +N CO +2N CO 2 , and

(D) Likewise,

A =N CO /N 1/2 O 2 (P/(1 ×N)) .

(E) where P' = P/1. Hence,

K = (N N 1/2 B CO O 2 /N CO 2 (P'/N) 1/2 ,

K B /K A =N O 2 /N CO 2 , and

(F)

Since 2a = 2 N O 2 +N CO +2N O 2 (K A /K B ), N O 2 = (2a – N CO )/(2(K A /K B +1)), and

(H) N=N CO + (2a – N CO )/(2(K A /K B + 1)) + (K A /K B ) (2a – N CO )/(2(K A /K B + 1))

= (2 a + N CO )/2. (I) Therefore, Eq. (D) becomes K A = (N

CO /((2a – N CO )/(2(K A /K B + 1))) ) (2 P'/(2 a + N CO )) 1/2 , or

A /(K A /K B + 1) =N CO (4 P'/(4a –N CO )) , i.e.,

(J) N 1/2

K 2 K /(K +K )=4N 2 P'/(4a 2 A 2 B B A CO –N CO ),

CO (T) = 2 a/(1 + 4 (P'/K A ) (1/K A +1/K B )) , and

(K)

CO (T) = N CO /N = 2/(1 + (1 + 4 (P'/K A ) (1/K A + 1/K B )) )). (L)

X CO 2 can be similarly expressed. The required carbon atom input is

(M) where the equality applies to the minimum carbon input required for achieving

N C,in ≥N CO +N CO 2 ,

chemical equilibrium. A mathematical expression for the minimum carbon input is N C,in,min /a = [{(K A /K B )+1}/{1+4(1/K B +1/K A )P'/K A } /2 + 2(K A /K B )]/(2+K B /K A ).