HW # 6 Chemical Equilibrium

Transcription

1 HW # 6 Chemical Equilibrium roblem 1: Noggle 6.1 ropane is commonly used for heating and in applications like gas grills. Calculate the maximum temperature of propane burning in air from its heat of combustion: kj/mole. The reaction (assume 2% O2 and 8% N2 in air): C3H8(g) + 5O2(g) + 2N2(g) = 3CO2(g) + 4H2O(g) + 2N2(g) ΔH = kj/mole when products are at 25C. Use the heat to raise the temperature of the products. Heat capacities to heat CpCO2( T) T T gaseous products (per mole): T 2 CpN2( T) T T Noggle page 133 T 2 CpH2O( T) T T T 2 ΔH vap ( ) 1 Section 6.2 page 133 gives ΔHvap for vaporizing water Given guess: T2 2 ΔH vap = T2 298 ( CpCO2()3 t + CpN2()2 t + CpH2O()4 t ) dt ΔH vap Tad Find( T2) Tad = roblem 2: ( Noggle 6.5) Use the table to calculate the fugacity coefficients of CO 2 at K. Make a graph of the fugacity coef. versus and report the values for 2, 4, 6, 8 and 1 atm. The definition of the fugacity coefficient for a gas is given by

2 bar atm data Φ f f f 1bar the expression of Φ in terms of Z and is : ln f bar Z ( ) d ( ) ( ) 1 data2 T atm Z data2 T For the fugacity coefficient, you will need a smooth function of Z(). Use of the spline fit. vs cspline(, Z) fit1( p) interp( vs,, Z, p) p atm, 5atm.. 1atm 1.5 Z fit1( p) p, atm atm With the expression (Z()-1)/ you will run into a division by zero unless the integrand can be changed to a constant limit of (Z-1)/p as p---->. New fit (Z-1)/ versus. i.. 13 integrand fit: vs Z i 1 y i i cspline(, y) fit2( p) interp( vs,, y, p) y i atm fit2( p) atm i p, atm atm

3 initial press:.5 atm.5 atm atm atm equil moles.5-4x.5-x 2x 2x Total moles at equil: 1-x equil mole fr.: fhcl=(.5-4x)/(1-x) fo2=(.5-x)/(1-x) fcl2=2x/(1-x) fh2o=2x/(1-x) Now, this can be integrated without a singularity at zero pressure. 1 1 atm lnφ( 1) fit2( p) dp lnφ 1 φ( 1) e ( ) 1 φ( p).5 rr 2atm, 4atm.. 1atm 5 1 p atm φ( rr) = roblem 3: Noggle 6.8 For the equilibrium 4 HCl(g) + O 2 (g) = 2 Cl 2 (g) + 2 H 2 O(g) calculate the mole fraction and partial pressure of Cl 2 for a (total) equilibrium pressure of 6, 12 and 22 atom when a 1:1 mixture of the reactants is used. This is the Deacon process which was once used for the commercial production of chlorine. Use Kp = atm-1 for Solution set-up: Let x be the "extent of reaction", then 723K. 4 HCl(g) + O 2 (g) = 2 Cl 2 (g) + 2 H 2 O(g) initial press:.5 atm.5 atm atm atm equil moles.5-4x.5-x 2x 2x Total moles at equil: 1-x equil mole fr.: fhcl=(.5-4x)/(1-x) fo2=(.5-x)/(1-x) fcl2=2x/(1-x) fh2o=2x/(1-x) Solution set-up: Let x be the "extent of reaction", then 4 HCl(g) + O 2 (g) = 2 Cl 2 (g) + 2 H 2 O(g)

4 22atm guess extent of rx: guess extent of rx: K( x) 2x 2 4 (.5 4x) x.12 solve for extent of rx at equilibrium: solve for extent of rx at equilibrium: extent =.111 set x extent set Calculate mole fractions: Calculate mole fractions:.5 4x fhcl fhcl =.64.5 x fo2 fo2 =.438 2x fcl2 fcl2 =.249 2x fh2o fh2o =.249 2x 2 (.5 x) simplify: simplify: K( x) sensitive when x=.5 = sensitive when x=.5 = x 4 ( ) (.5 4x) 4 (.5 x) ( ) extent root K( x) 23.14atm 1, x Calculate artial ressures: Calculate artial ressures: HCl fhcl HCl = 1.417atm O2 fo2 O2 = atm Cl2 fcl2 Cl2 = atm H2O fh2o H2O = atm check: check: fhcl + fo2 + fcl2 + fh2o = 1 Noggle 6.1 For the Fischer-Tropsch reaction at 7K (Example 6.11), calculate the mass of methane formed if 5 moles of CO, 5 moles of water and 5 moles of CO 2 (no H 2 ) are reacted at 5 bar. What percent of the total carbon present is converted to methane? This is an example of multiple equilibria. In this case, the reactions are CO + 3 H 2 = CH 4 + H 2 O K CO init -x H2 init -3x CH4 init +x H2O init +x and CO + H 2 O = H 2 + CO 2 K COinit-y H2Oinit-y H2init+y CO2init+y where x is the extent of reaction in the first reaction and y is the extent of reaction in the second.

5 Initial: COinit 5 H2Oinit 5 CO2init 5 H2init CH4init Total pressure: 5 bar bar 1 moles at equilibrium: n CO ( x1, y1) COinit x1 y1 n H2O ( x1, y1) H2Oinit + x1 y1 n H2 ( x1, y1) H2init 3x1 + y1 n CO2 ( x1, y1) CO2init + y1 n CH4 ( x1, y1) CH4init + x1 n total ( x1, y1) n CO ( x1, y1) + n H2O ( x1, y1) + n H2 ( x1, y1) + n CO2 ( x1, y1) + n CH4 ( x1, y1) guess x and y: x1.5 y1 2.1 tolerance: TOL 1 5 Solve using a Given block and the Find function: Given mass balance for carbon: n CO ( x1, y1) + n CO2 ( x1, y1) + n CH4 ( x1, y1) COinit + CO2init + CH4init extent and total moles are positive: x1 y1 x1 > y1 > n total equilibrium constants with fractions removed (simplifies solution): K 1 2 n CO ( x1, y1) n H2 ( x1, y1) 3 K 2 ( n CO ( x1, y1) n H2O ( x1, y1) ) Find( x1, y1) This is the end of the solve block Solution: x1 = y1 = ( x1, y1) > n CH4 ( x1, y1) n H2O ( x1, y1) n total ( x1, y1) 2 n H2 ( x1, y1) n CO2 ( x1, y1) ERR = moles of methane: n CH4 ( x1, y1) = mass: gm M ( ) mole mass n CH4 ( x1, y1) molem mass = gm n CH4 ( x1, y1) percent COinit + CO2init + CH4init percent = % Alternative way to solve this is: n CO ( x2, y2) 5 x2 y2 n H2 ( x2, y2) 5 3x2 + y2

6 n CH4 ( x2, y2) x2 n H2O ( x2, y2) 5 + x2 y2 n total ( x2, y2) 23.4 The substitute these terms in the two expressions below x2.5 y2 2.1 Given K 1 2 ( 5 x2 y2) ( 5 3x2 + y2) 3 x2( 5 + x2 y2) n total ( x2, y2) 2 K 2 ( 5 x2 y2) ( 5 + x2 y2) ( 5 3x2 + y2) ( 5 x2 y2) x2 y2 Find( x2, y2) x2 = and y2 = 3.765