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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)

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

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