Answer this one:

DAVID :) Problem Set CN 12 9. Consider the reaction: H218) + 12(5) 2HI(g) If the initial concentrations of H2 and l, are 0.500 M, what are the equilibrium concentrations of each species? Kc = 54.3 at 35.C " Answer: [H2] = 0.107 M, [/2] = 0.107 M, THI - 0.786 M KC: 54:3 at 35 C COMPUTATION OF VALUE Ha (q)/+ 12(g) 2HI (9) Hz Iz 2 HI initial concentration I(M) 0-500 0.500 0 0:500 0-300 Change in concentration C (M) 7X - X $2x C -0-395 M -0-393 M 0:386 M equilibrium concentration E (ML/0 500 -x) Q.500 -x m 1-54-0 395 0:300 5 383 2(0 393 KC = [2HI]] [ H2] [In ] 7.37(0.500-X) = 2x 3-69-737X = 2X (0.500-x) 369 - 957x 9-37 957 137 - 2x 0.500 X= 0-39310 Consider the reaction Calculate [[ ] and ] at equilibrium at 600 C when initial concentration of I. 15 0.20 M. Ke 2.94 x 1010 Answer: [L.] - 0.20 M, - 7.6 x10 M