Consider the following reaction: 2A + 3 B_2 A_2B_6 The initial concentrations of

Question

Consider the following reaction: 2A + 3 B_2 A_2B_6 The initial concentrations of A and B2 are 0.290 M and 0.550 M, respectively, with no A_2B_6 initially present. When the reaction comes to equilibrium, the concentration of A_2B_6 is found to be 4.50 times 10^-2 M. What is the equilibrium concentration of A? Round your answer to 3 significant figures. What is the equilibrium concentration of B_2? Round your answer to 3 significant figures. What is the value of the equilibrium constant? Round you answer to 3 significant figures.

Explanation / Answer

           2 A       +         3 B2 ---->   A2B6
initially        0.29                  0.55          0
at equi       (0.29-x/2)       (0.55-x/3)    x
Kc = [A2B6] / [A]^2[B2]^3
Kc = x / ((0.29-x/2)^2 * (0.55-x/3)^3)
Kc = (4.5*10^-2)/((0.29-(4.5*10^-2/2))^2*(0.55-(4.5*10^-2/3))^3)
Kc = 4.107
equilibium concentration of A = (0.29-(4.5*10^-2/2))^2 = 0.0715
equilibium concentration of B = (0.55-(4.5*10^-2/3))^3 = 0.153

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