Predict and calculate the effect of concentration changes on an equilibrium system.

Question

Some (CH₃)₂CHOH is allowed to dissociate into (CH₃)₂CO and H₂ at 452 K. At equilibrium, [(CH₃)₂CHOH] = 0.244 M, and [(CH₃)₂CO] = [H₂] = 5.41×10^-2 M. Additional (CH₃)₂CO is added so that [(CH₃)₂CO]_new = 9.04×10^-2 M and the system is allowed to once again reach equilibrium.

1. (CH₃)₂CHOH(g) (CH₃)₂CO(g) + H₂(g)   K = 1.20×10^-2 at 452 K

(a) In which direction will the reaction proceed to reach equilibrium? to the right or to the left

(b) What are the new concentrations of reactants and products after the system reaches equilibrium?

[(CH3)2CHOH]=

[(CH3)2CO]=

[H2]=

Answer 1

(CH3)2CHOH(g) <==> (CH3)2CO(g) + H2(g) ... K = 1.20x10^-2

According to Le Chatelier's principle, adding (CH3)2CO(g) to the system @ equilibrium will shift the reaction to the left, towards the reactant side.

(CH3)2CHOH(g) <==> (CH3)2CO(g) + H2(g)

0.244............................0.0904............0.0541.........Initial

+x.................................-x......................-x...............Change

0.244+x......................0.0904-x..........0.0541-x.......Equilibrium

K = 1.20x10^-2 = (0.0904-x)(0.0541-x) / (0.244+x)

Solving for x, and you should check this calculation, we

x = 0.0170

Thus, @ equilibrium we should have...

[(CH3)2CHOH] = 0.244 + 0.0170 = 0.261 M

[(CH3)2CO] = 0.0904 - 0.0170 = 0.0734 M

[H2] = 0.0541 + 0.0170 = 0.0711 M