At a particular temperature, Kc = 54 for the reaction H2(g) + I2 (g) <=> 2 HI(g). One mole of HI is placed in a 3.0-L container. What would be the equilibrium concentration of HI?
H2 + I2 = 2HI I 1.0/5.0 1.0/5.0 0 C -x -x 2x E 0.2-x 0.2-x 2x 54= [HI]^2/[H2]*[I] 54=2x^2/(0.2-x)(0.2-x) need to solve for x then would I take square root of both sides to get rid of the squares? Then I get stuck at 7.348= 4x/0.2-x I dont know where else to go from here and i...
At a particular temperature, Kc = 54 for the reaction If 1.0 mole of H2 and 1.0 mole of I2 are placed in a 5.0 L container, what would be the equilibrium concentration of HI?
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