Question

4. Consider the following reaction at equilibrium CO(g) + H2O(g) + CO2(g) + H2(g) 2.50 mole of CO(g) and 2.50 mole of H2O(g) gas at 588 K are mixed in a 10.00 L container. (Kc = 31.4 at 588 K) Calculate the concentration of CO(g), H2O(g), CO (g), and H.(g) at equilibrium.

Answer 1

Initially moles of moles of co= H2O = 2.50 2.50 10 Volume of container & LOL Malarity of cos 2050 0.25 Molarity of H20 = dose - 0.25% Colg) + H2O(g) 2 congs + Hargs o t=o t=teg 0.25 10.25-2) 0.25 (0.25-21) Kc = [602) [H₂ 1 [co) (HD). (0.25-x) (0.25-27

By taking square rost on side both 5.60 = x _0025-2 lou = 6oGa =) a= 0.212 M At equilibrium- [col = 025-0.212 = I H2o] : 0.25-00212= I co2l = 0.212 M [ H2) = 0. 212M 0.03819 0.038M

Colg) + H2g cog) + H206) ceives [col= 2.50 20:25-M [ H20]2_2:59 20-25 09 [ CO2] = 5 = 0.5M [H2) = s = 0.8m = = a - [102] [H2) [co] fH2O) 10.5) loos) 10.25) 10.25) = 024. As given Ke= 31.4 - 0 <kc so in Reaction forward will move eleirection. - A Colg + H2O1g) 2 t-o 0.25 0.25 tateg (0025-0) (0.25-1)_ coulgo + H2(g) 050.5 1005+x)_ 1005+) in Icz (0.5+0) 10.5+1) 10.259->) (0.25->) = 3104 =) 3104= 10.5+212 10.25-122 By taking square root on both side I 5.6 = 0.5t I only positive value wi we will 025-2 which e accepted because by taky negative a will be negative от acceptable ra. egative

By taking square rost on side both 5.60 = x _0025-2 lou = 6oGa =) a= 0.212 M At equilibrium- [col = 025-0.212 = I H2o] : 0.25-00212= I co2l = 0.212 M [ H2) = 0. 212M 0.03819 0.038M

Please give a thumbs-up.

Do comment if you have any doubts

Thank you

Best of luck