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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