we know that
concentration = moles / volume (L)
so
initially
[HI] = 0.66 / 2 = 0.33
now
2HI ----> H2 + I2
using ICE table
at equilibrium
[HI] = 0.33 -2x
[H2] = x
[I2] = x
now
Kc = [H2] [I2] /[HI]^2
so
0.0156 = [x] [x] / [0.33-2x]^2
0.0156 = [x]^2 / [0.33-2x]^2
solving we get
x = 0.033
so
33)
at equilibrium
[H2]= x = 0.033 M
option E) 0.033 M
34)
[I2] = x = 0.033 M
option E) 0.033 M
35)
[HI] = 0.33 - 2x
[HI] = 0.33 - ( 2 * 0.033)
[HI] = 0.264 M
option A) 0.264 M
Use the information below to answer questions 33-35. Hydrogen iodide decomposes according to the equation: 2HI(g)...
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