Please show work. Thanks ! part a. The system: A(g) + C(g) 2B(g) Kc = 1.00 x 10-

Question

Please show work. Thanks !

part a. The system: A(g) + C(g) 2B(g) Kc = 1.00 x 10-4 initially started with [B] = 1.20M . The equilibrium concentration of B is:

A.0.109 B.0.00600 C.0.0109

part b. For the reaction A(g) + 2B(g) C(g) ;  Kc = 10   Which direction will it proceed if [A] = 0.100, [B]= 0.100 , and [C] = 0.100 is found before equilibrium is established? Left, right or no change

part c.

For the system: 2CO(g) + O2(g)   2CO2(g) ; increasing the volume of the container will shift the equilibrium: Left or right ? or stay the same ?

Explanation / Answer

a. 2B <---> A + C

I 1.20 0 0

C -x +x +x

E 1.20-x +x +x

Kc = 1.00 x 10-4 = [A][C]/ [B]2= (x)*(x)/(1.20-x) hence, x= 0.0109 answer C.

part b.  if [A] = 0.100, [B]= 0.100 , and [C] = 0.100 is found before equilibrium is established

then at that point using formula for Kc, it gives Kc = 100. So, the reaction will proceed to the right.

part C. Total number of atoms are decreasing towards the CO2 side. Now increasing the volume of the container means decreasing the pressure, hence according to Le Chatelier's principle, to nullify the effect of that decreased pressure the system will tend to increase the number of molecules or atoms i.e. the equilibrium will move towards the left hand side.

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