Test Index

NCERT Class XI Chemistry Equilibrium Solutions

© examsnet.com
Question : 23
Total: 73
At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55% CO by mass
C(s)+CO2(g) ⇌ 2CO(g)
Calculate Kc for this reaction at the above temperature.
Solution:  
Let the total mass of the mixture of CO and CO2 is 100 g, then
CO = 90.55 g and CO2 = 100 – 90.55 = 9.45 g
Moles of CO =
90.55
28
 = 3.234 ; Moles of CO2 =
9.45
44
 = 0.215
Mole of fraction of CO =
3.234
3.234+0.215
 = 0.938
Mole fraction of CO2 =
0.215
3.234+0.215
 = 0.062
∴ pCO = mole fraction × total pressure = 0.938 × 1 atm = 0.938 atm
pCO2 = 0.062 × 1 atm = 0.062 atm
Kp for the reaction C(s)+CO2(g) ⇌ 2CO(g)
Kp =
pCO2
PCO2
 =
(0.938)2
0.062
 = 14.19
Now Δng = 2 - 1 = 1 , Kp = Kc(RT)Δng
or Kc =
Kp
RT
 =
14.19
0.0821×1127
 = 0.153
© examsnet.com
Go to Question:
Prev Question
Next Question