7.23 At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with soild carbon has 90.55% CO by mass

C (s) + CO2 (g) 2CO (g)

Calculate Kc for this reaction at the above temperature.

Let the total mass of the gaseous mixture be 100 g.
Mass of CO = 90.55 g
And, mass ofCO2= (100 – 90.55) = 9.45 g

Now, the number of moles of COnco=90.5528=3.234mol

 Number of moles of CO2,nco2=9.4544=0.215mol

The partial pressure of CO, 

 

pco=nconco+nco2×ptotal
=3.2343.234+0.215×1
=0.938atm

The partial pressure of CO2,

 

pco2=nco2nco+nco2×ptotal
=0.2153.234+0.215×1
=0.062atm

Therefore Kp=

=CO2CO2
=0.93820.062
=14.19

For the given reaction,

n=2-1=1
Weknowthat,
Kp=KcRTn
14.19=Kc0.082×11271
Kc=14.190.082×1127
=0.154approx.