Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) + I
ID: 483807 • Letter: K
Question
Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) + I2(g) <> 2HI(g) is 54.3 at 430c. what will be the concentrations of all species at equilibrium if we start with 0.375 M of H2 and I2? Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) + I2(g) <> 2HI(g) is 54.3 at 430c. what will be the concentrations of all species at equilibrium if we start with 0.375 M of H2 and I2? Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) + I2(g) <> 2HI(g) is 54.3 at 430c. what will be the concentrations of all species at equilibrium if we start with 0.375 M of H2 and I2?Explanation / Answer
first construct ICE table
H2 (g) + I2 (g) <-----------> 2HI (g)
I 0.375M 0.375 M 0
C -x -x +2x
E 0.375-x 0.375-x +2x
Kc = [HI]2 / [H2] [I2]
54.3 = [2x]2 / [0.375-x] [0.375-x]
54.3 = 4x2 / 0.140625 + x2 - 0.75x
54.3x2 - 40.725x + 7.636 = 4x2
50.3x2 - 40.725x + 7.636 = 0
solve the quadratic equation
you will have a two values
which are x = 0.5147 and x = 0.295
0.5147 is more than intial concentration so we can neglect that
valide solution is x = 0.295
now
equilibrium concentrations of
[H2] = 0.375 - x = 0.375 - 0.295 = 0.08 M
[I2] = 0.375 - x = 0.375 - 0.295 = 0.08 M
[HI] = 2x = 2 x 0.295 = 0.59 M
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