Question

5) During an where the following reaction came to equilibrium. etion K s49.5 at the temperature of the experiment. were the equilibrium concentrations of H2, Ia, and HI? H2(g) + 12(g)与2H1(g) a) What [12] = [H11-
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Answer #1

H2(g) + I2(g) <------> 2HI(g)

Kc = [ HI ]^2/[ H2 ] [ I2 ] = 49.5

Initial concentration

[ H2 ] = (0.986mol/1.95L)×1L = 0.5056M

[ I2 ] = ( 0.493mol/1.95L)×1L = 0.2528M

[ HI ] = 0

concentration at equillibrium

[ H2 ] = 0.5056 - x

[ I2 ] = 0.2528 - x

[ HI ] = 2x

Therefore,

(2x)^2 /(0.5056 - x ) × ( 0.2528 - x) = 49.5

4x^2/(0.1278 -0.5056x - 0.2528x + x^2) = 49.5

4x^2/(x^2 - 0.7584x + 0.1278 ) = 49.5

45.5x^2 - 37.5408x + 6.3261=0

x = 0.2361

Therefore,

Equillibrium concentrations are

[ H2 ] = 0.5056 - 0.2361= 0.2695M

[ I2 ] = 0.2528 - 0.2361 = 0.0167M

[ HI ] = 2×0.2361 = 0.4722M

  

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