Question

Consider the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) Kc=102 at 500 K A reaction mixture initially contains 0.150 MCO and 0.150 MH2O.

Part A

What will be the equilibrium concentration of CO? Express the concentration in molarity to two significant figures.

Answer 1

Given reaction is CO_g + H₂O_g CO_2g + H_2g

Equilibrium constant K_c = 102

Lets write the ice table:    CO_g + H₂O_g CO_2g + H_2g

Initial concentration 0.15 0.15 0 0

Change in concentration -x -x +x +x

Equilibrium concentration 0.15-x 0.15-x x x

Now K_c = [CO₂][H₂] / [CO][H₂O] = x*x/(0.15-x )*(0.15-x ) = 102

=> x²/(0.15-x )² =102

=> x/(0.15-x) = = 10.1

=> x= (0.15-x)*10.1

=> x = 1.515 - 10.1x

=> 11.1x = 1.515

=> x = 0.136

Therefore equilibrium concentration of CO = 0.15 - 0.136 = 0.014M

Hope this helps.