anonymous 4 years ago 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)

1. anonymous

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}$

2. anonymous

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