24. Suppose that 0.350 mol of A and 0.520 mol of B are placed a 1.50 L container and the following hypothetical equilibrium is established. If the equilibrium amount of C is 0.150 mol, what is the equilibrium constant for this reaction?
A(g) +2B(g) ---3C(g)

Question

24.	Suppose that 0.350 mol of A and 0.520 mol of B are placed a 1.50 L container and the following hypothetical equilibrium is established. If the equilibrium amount of C is 0.150 mol, what is the equilibrium constant for this reaction?
A(g) +2B(g) --->3C(g)

xishem · Answer

First, you're going to want to set up an ICE chart:

$$[A]_0=\frac{0.350mol}{1.5L}=0.23M$$$$[2B]_0=\frac{0.520mol}{1.5L}=0.35M$$$$\Delta [A]=\frac{-1}{3}(0.10M)=-0.033M$$$$\Delta [2B]=\frac{-2}{3}(0.10M)=-0.067M$$
Concentration (M)  [A]               [2B]           [3C]
Initial                    0.23           0.35             0
Change               -0.033      -0.067         0.100
Equilibrium           0.200        0.280         0.100

Then, we can go ahead and plug these in to our K_c equation:$$K_c=\frac{[C]^3}{[A][B]^2}=\frac{(0.100)^3}{(0.200)(0.280)^2}=6.38*10^{-2}$$$$K_c=6.38*10^{-2}$$

xishem · Answer

I forgot to write down a calculation which may cause confusion. I just want to clarify:$$[3C]_E=\frac{0.150mol}{1.5L}=0.100$$