Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
Question
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Chapter 14, Problem 14.16P

(a)

Interpretation Introduction

Interpretation:

The fractional conversion of propane for the given reaction is to be calculated at 625 K .

Concept introduction:

For a single reaction system, the final moles of each of the components present, can be estimated by the equation:

  ni=nio+viξ      .......(1)

Here, ni is the final moles of the component i, nio is the initial moles of the component i, vi is the stoichiometric coefficient of the component i in the reaction and ξ is the extent of the reaction. vi is taken as negative for reactants and positive for products.

Mole fraction (yi) of any component is given by:

  yi=nini      .......(2)

Here, ni is the moles of component i and ni is the total moles of all the components present in the mixture.

Equilibrium constant of this reaction from equation 14.28 can be written as:

  K=i(yi)νi( P 0 P)ν      .......(3)

Where, yi is the mole fraction of component i, P is the pressure of the reaction, and P0 is the standard pressure.

Gibb’s free energy in terms of equilibrium constant is written as:

  lnK=ΔG0RT      .......(4)

Also, Gibbs free energy is calculated using heat of reaction from the equation given as 14.11b .

  ΔG0=ΔH00TT0(ΔH00ΔG00)+RT0TΔCP0RdTRTT0TΔCP0RdTT      .......(5)

Here, T0TΔCP0RdT and T0TΔCP0RdTT are defines as:

   T 0T Δ C P 0 RdT=(ΔA)T0(τ1)+ΔB2T02(τ21)+ΔC3T03(τ31)+ΔDT0(τ1τ)T0TΔ C P 0RdTT=ΔAlnτ+(τ1)[ΔBT0+(τ+12)(ΔCT02+ΔD τ 2 T 0 2)]      .......(6)

Where, τ=TT0

(a)

Expert Solution
Check Mark

Answer to Problem 14.16P

The fractional conversion of propane for the given reaction at 625 K is 0.777 .

Explanation of Solution

Given information:

The cracking reaction of propane is:

  C3H8(g)C2H4(g)+CH4(g)

The equilibrium pressure given for this reaction is 1 bar .

The equilibrium conversion at 300 K is negligible but at temperatures above 500 K, it becomes appreciable.

The equilibrium temperature for this reaction is taken as 625 K .

From Table C.4 the standard heat of reaction and Gibb’s free energy for this reaction is:

  ΔG2980=42290J/molΔH2980=82670 J/mol

From Table C.1 the coefficients for the heat capacity of the component gases are given as:

    Substance   A   B   C   D
      C3H8   1.213   28.785×103   8.824×106   __
      C2H4   1.424   14.394×103   4.392×106   __
      CH4   1.702   9.081×103   2.164×106   __

ΔA, ΔB, ΔC, and ΔD from above given values can be calculated as:

  ΔA=AC2H4+A CH4AC3H8=1.424+1.7021.213=1.913ΔB=BC2H4+B CH4BC3H8=(14.394+9.08128.785)×103=5.31×103ΔC=CC2H4+C CH4CC3H8=(4.392+( 2.164)( 8.824))×106=2.268×106ΔD=DC2H4+D CH4DC3H8=0+00=0

Now, use equations set (6) to evaluate the values of T0TΔCP0RdT and T0TΔCP0RdTT at equilibrium temperature of 625 K as:

   τ= T T 0 = 625 K 298.15 K =2.096

   T 0 T Δ C P 0 R dT =( ΔA ) T 0 ( τ1 )+ ΔB 2 T 0 2 ( τ 2 1 )+ ΔC 3 T 0 3 ( τ 3 1 )+ ΔD T 0 ( τ1 τ )

   =( ( 1.913 )( 298.15 )( 2.0961 )+ ( 5.31× 10 3 ) 2 ( 298.15 ) 2 ( ( 2.096 ) 2 1 ) + ( 2.268× 10 6 ) 3 ( 298.15 ) 3 ( ( 2.096 ) 3 1 )+ ( 0 ) 298.15 ( 2.0961 2.096 ) )

   =11.259

   T 0 T Δ C P 0 R dT T =ΔAlnτ+( τ1 )[ ΔB T 0 +( τ+1 2 )( ΔC T 0 2 + ΔD τ 2 T 0 2 ) ]

   =( 1.913 )ln( 2.096 )+( 2.0961 )[ ( 5.31× 10 3 )( 298.15 )+( 2.096+1 2 )( ( 2.268× 10 6 ) ( 298.15 ) 2 + 0 ( 2.096 ) 2 ( 298.15 ) 2 ) ]

   =0.0226

Now, use equation (5) along with the above calculated values to get the value of ΔG0 as:

  ΔG0=ΔH00τ(ΔH00ΔG00)+R T 0T Δ C P 0 RdTRTT0TΔ C P 0RdTT=(82670)(2.096)(8267042290)+(8.314)(11.259)(8.314)(625)(0.0226)=2177.5 J/mol

Use this calculated value of ΔG0 to calculate the equilibrium constant of the reaction at 625 K using equation (4) as:

  lnK=ΔG0RTK=exp( ( 2177.5 ) ( 8.314 )( 625 ))K=1.521

For the given cracking reaction, let the extent of the reaction be ξ .

  C3H8(g)C2H4(g)+CH4(g)    1          0              0  1ξ           ξ           ξ

The individual stoichiometric coefficients for all the components in this reaction is:

  νC3H8=1νC2H4=+1ν CH4=+1

The overall stoichiometric coefficient for this reaction is,

  ν=1+1(1)=1

Using equation (1), write the expressions for final moles of all the components present in the products as gases.

  nC3H8=n0( C 3 H 8 )+vC3H8ξ=1ξnC2H4=n0( C 2 H 4 )+vC2H4ξ=ξn CH4=n0( CH 4 )+v CH4ξ=ξ

Total moles of the products will be:

  n=nC3H8+nC2H4+n CH4=1ξ+ξ+ξ=1+ξ

Using equation (2) to write the mole fraction of all the species as:

  yC3H8=1ξ1+ξyC2H4=ξ1+ξy CH4=ξ1+ξ

Now, use equation (3) and the calculated value of the equilibrium constant and calculate ξ . The standard pressure is taken as 1 bar .

  K=i( y i)νi( P 0 P)ν1.521=( y C 2 H 4 )1( y CH 4 )1( y C 3 H 8 )1( 1 1)(1)1.521=( ξ 1+ξ )( ξ 1+ξ )( 1ξ 1+ξ )ξ=0.777

Therefore, the fractional conversion of propane is 0.777 .

(b)

Interpretation Introduction

Interpretation:

The temperature for the given fractional conversion of propane at 1 bar is to be determined.

Concept introduction:

For a single reaction system, the final moles of each of the components present, can be estimated by the equation:

  ni=nio+viξ      .......(1)

Here, ni is the final moles of the component i, nio is the initial moles of the component i, vi is the stoichiometric coefficient of the component i in the reaction and ξ is the extent of the reaction. vi is taken as negative for reactants and positive for products.

Mole fraction (yi) of any component is given by:

  yi=nini      .......(2)

Here, ni is the moles of component i and ni is the total moles of all the components present in the mixture.

Equilibrium constant of this reaction from equation 14.28 can be written as:

  K=i(yi)νi( P 0 P)ν      .......(3)

Where, yi is the mole fraction of component i, P is the pressure of the reaction, and P0 is the standard pressure.

Gibb’s free energy in terms of equilibrium constant is written as:

  lnK=ΔG0RT      .......(4)

Also, Gibbs free energy is calculated using heat of reaction from the equation given as 14.11b .

  ΔG0=ΔH00TT0(ΔH00ΔG00)+RT0TΔCP0RdTRTT0TΔCP0RdTT      .......(5)

Here, T0TΔCP0RdT and T0TΔCP0RdTT are defines as:

   T 0T Δ C P 0 RdT=(ΔA)T0(τ1)+ΔB2T02(τ21)+ΔC3T03(τ31)+ΔDT0(τ1τ)T0TΔ C P 0RdTT=ΔAlnτ+(τ1)[ΔBT0+(τ+12)(ΔCT02+ΔD τ 2 T 0 2)]      .......(6)

Where, τ=TT0

(b)

Expert Solution
Check Mark

Answer to Problem 14.16P

The temperature for 85% fractional conversion of propane at 1 bar is 647 K .

Explanation of Solution

Given information:

The cracking reaction of propane is:

  C3H8(g)C2H4(g)+CH4(g)

The equilibrium pressure given for this reaction is 1 bar .

The equilibrium conversion at 300 K is negligible but at temperatures above 500 K, it becomes appreciable.

The fractional conversion at equilibrium is 85% .

The initial and final pressure of the system is kept at 1 bar, so, P=P0=1 bar .

Fractional conversion of propane is given as 85%, so, ξ=0.85 for the given reaction.

The final mole fraction of all the species will be:

  yC3H8=1ξ1+ξ=10.851+0.85=0.0811yC2H4=ξ1+ξ=0.851+0.85=0.4595y CH4=ξ1+ξ=0.851+0.85=0.4595

Now, use equation (3) to calculate the equilibrium constant of this reaction as:

  K=i( y i)νi( P 0 P)νK=( y C 2 H 4 )1( y CH 4 )1( y C 3 H 8 )1( 1 1)(1)K=( 0.4595)( 0.4595)( 0.0811)K=2.604

From Table C.4 the standard heat of reaction and Gibb’s free energy for this reaction is:

  ΔG2980=42290J/molΔH2980=82670 J/mol

From part (a), the calculated values of ΔA, ΔB, ΔC, and ΔD are:

  ΔA=1.913ΔB=5.31×103ΔC=2.268×106ΔD=0

Make an initial guess for the equilibrium temperature at 85% conversion. Basing the guess on t he given information and the result of part (s), let the equilibrium temperature be 647K .

Now, use equations set (6) to evaluate the values of T0TΔCP0RdT and T0TΔCP0RdTT at this equilibrium temperature of 647 K as:

Now, use equation (5) along with the above calculated values to get the value of ΔG0 as:

  ΔG0=ΔH00τ(ΔH00ΔG00)+R T 0T Δ C P 0 RdTRTT0TΔ C P 0RdTT=(82670)(2.170)(8267042290)+(8.314)(23.318)(8.314)(625)(0.0036)=5167.2 J/mol

Use this calculated value of ΔG0 to calculate the equilibrium constant of the reaction at 647 K using equation (4) as:

  lnK=ΔG0RTK=exp( 5167.2 ( 8.314 )( 647 ))K=2.613

Since, the calculated value of K and the value calculated using the guessed value of T are almost same, the temperature initially guessed is the equilibrium temperature for 85% conversion of propane.

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Chapter 14 Solutions

Introduction to Chemical Engineering Thermodynamics

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