For CO(g) + 2H₂(g) CH₃OH(g) , Kc = 10 at 400 K. If 0.1 mol CO and 0.3 mol H₂ are in a 1 L vessel, what is [CH₃OH] at equ
Initial: [CO] = 0.1 M , [H₂] = 0.3 M , [CH₃OH] = 0 . Let x = [CH₃OH] , [CO] = 0.1 - x , [H₂] = 0.3 - 2x . Kc = ([CH₃OH]/[CO][H₂]²) = (x/(0.1 - x)(0.3 - 2x)²) = 10 . Solving iteratively, x ≈ 0.09 , 10 = (0.09/(0.01)(0.12)²) ≈ 625 (too high), adjust x ≈ 0.06 , (0.06/(0.04)(0.18)²) ≈ 46 (still high), x ≈ 0.03 , (0.03/(0.07)(0.24)²) ≈ 7.44 , close to 10.
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law