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 of CO2 = (100 - 90.55) = 9.45 g
Now, number of moles of CO, nco = 90.55/28 = 3.234 mol
Number of moles of CO2, nco2 = 9.45/44 = 0.215 mol
Partial pressure of CO,
For the given reaction,
Δn = 2 - 1 = 1
we know that,
Kp = Kc (RT)Δn
⇒ 14.19 = Kc (0.082 x 1127)1
⇒ Kc = 14.19 / 0.082 x 1127
= 0.154 (arrpox.)
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 23: At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with soild carbon has 9....
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