# Class 11 Chemistry Equilibrium: NCERT Solutions for Question 23

This page focuses on the detailed Equilibrium question answers for Class 11 Chemistry Equilibrium, addressing the question: '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.'. The solution provides a thorough breakdown of the question, highlighting key concepts and approaches to arrive at the correct answer. This easy-to-understand explanation will help students develop better problem-solving skills, reinforcing their understanding of the chapter and aiding in exam preparation.
Question 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 massC (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.)