Question

84. The equilibrium constant (K) for this reaction is 5.0 at a given temperature. CO(g) + H2O(g) + CO2(g) + H2(g) (a) On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of CO, 0.30 mol of water vapor, and 0.90 mol of H2 in a liter. How many moles of CO2 were there in the equilibrium mixture? (b) Maintaining the same temperature, additional H2 was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of CO, 0.30 mol of water vapor, and 1.2 mol of H2 in a liter. How many moles of CO2 were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established.

Answer 1

Reaction taking place; CO(g) + tho 0.20 0.30 co + H₂ - 0.90 0.20 mol mol mol "Ke= 5= (moles) [co] [Hz] [col[ho] Volume) 5= Ncor (6.90) (0.20 (0:30) co₂ = 0.33 moles 6 V . We have maintained same temp. & Ke will remain same Co + H2O - CO₂ + H2 1.2mol Ategin: 0.40 0.30 - mol mol Ke=ncoz (1.2) (6 TOMO) (0.30)

= 0.5 moks It is totally reasonable for Moz= 0.5 mules Because when in and case concentration of each molecule (except tho) has increased to certain amount. As we are removing H₂O( reactant) more and more egin will shift in Backward direction is to reactant sick to produce tho (Le-chattier principle) Thus at egin we will agair have 0.30 mol of the vapour. ww)