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//(14.1) Evaluate the equilibrium constant, expressed as log10K, for the reaction at (a) 298 K and (b) 2000 K. Compare with the value obtained from Table A-27.
//solution
//The reaction is CO + .5O2 ----> CO2
//part(a)
T = 298 //in kelvin
Rbar = 8.314 //universal gas constant in SI units
//from table A-25
hfbarCO2 = -393520 //in kj/kmol
hfbarCO = -110530 //in kj/kmol
hfbarO2 = 0 //in kj/kmol
deltahbarCO2 = 0 //in kj/kmol
deltahbarCO = 0 //in kj/kmol
deltahbarO2 = 0 //in kj/kmol
sbarCO2 = 213.69 //in kj/kmol.K
sbarCO = 197.54 //in kj/kmol.K
sbarO2 = 205.03 //in kj/kmol.K
deltaG = [hfbarCO2-hfbarCO-.5*hfbarO2] + [deltahbarCO2-deltahbarCO-.5*deltahbarO2] - T*(sbarCO2-sbarCO-.5*sbarO2)
lnK = -deltaG/(Rbar*T)
logK = (1/log(10))*lnK
//from table A-27
logKtable = 45.066
printf('part(a) the value of equilibrium constant expressed as log10K is: %f',logK)
printf('\nthe value of equilibrium constant expressed as log10K from table A-27 is: %f',logKtable)
//part(b)
T = 2000 //in kelvin
//from table A-23
hfbarCO2 = -393520 //in kj/kmol
hfbarCO = -110530 //in kj/kmol
hfbarO2 = 0 //in kj/kmol
deltahbarCO2 = 100804-9364 //in kj/kmol
deltahbarCO = 65408 - 8669 //in kj/kmol
deltahbarO2 = 67881 - 8682 //in kj/kmol
sbarCO2 = 309.210 //in kj/kmol.K
sbarCO = 258.6 //in kj/kmol.K
sbarO2 = 268.655 //in kj/kmol.K
deltaG = [hfbarCO2-hfbarCO-.5*hfbarO2] + [deltahbarCO2-deltahbarCO-.5*deltahbarO2] - T*(sbarCO2-sbarCO-.5*sbarO2)
lnK = -deltaG/(Rbar*T)
logK = (1/log(10))*lnK
//from table A-27
logKtable = 2.884
printf('\n\npart(b) the value of equilibrium constant expressed as log10K is: %f',logK)
printf('\nthe value of equilibrium constant expressed as log10K from table A-27 is: %f',logKtable)
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