Chemistry
posted by Sandra .
A 100 mL container is filled with 0.01 mol of A2B3. At equilibrium, it also contains 0.006 mol of B. Find the value of K.
A2B3(g) > 2A(g) +3B(g)

........A2B3 ==> 2A + 3B
I......0.01.......0....0
C.......x........2x...3x
E....0.01x.......2x...3x
The problem tells us that 3x = 0.006 = mols B; therefore, x = 0.006/3 = 0.002
Equilibrium mols are as follows:
A2B3 = 0.010.002 = ?
A = 2*0.002 = ?
B = 0.006
These are mols. You must convert to molarity. Do that by M = mols/L. The volume is given as 0.1L. Then substitute concns into the Keq expression and solve for Kc.