Question

A mixture containing an initial concentration of 0.1614 M for H2 and 0.1440 M for I2...

A mixture containing an initial concentration of 0.1614 M for H2 and 0.1440 M for I2 is allowed to come to equilibrium (see reaction below). What must be the equilibrium concentration of HI?

H2(g) + I2(g) ↔ 2HI(g) Kc = 48.7000
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Answer #1

H2(g) + I2(g)      2HI (g)

initial concentration 0.1614   0.1440 0

equillibrium concentration 0.1614 - x    0.1440 -x 2x

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

48.7 = [2x]2 / [ 0.1614 - x   ] [0.1440 -x    ] = 4x2 / 0.02324 - 0.1614x - 0.1440x - x2

48.7 = 4x2 / 0.02324 -0.3054 -x2

48.7 ( 0.02324 -0.3054 -x2 ) = 4x2

1.1317 - 14.87x - 48.7x2 =  4x2

52.7x2 +  14.87x -   1.1317 = 0 ( using the quadratic equation we solve it )

x = -b +_ ( b2 - 4*a*c) /2a

= - 14.87 +_ ( 14.872 - 4*52.7*-1.1317) / 2 * 52.7

= - 14.87 +_ ( 221.12 + 238.56 ) / 105.4

=  - 14.87 +_ 459.68  / 105.4

=   - 14.87 + 21.44 /105.4

= 6.57 / 105.4 = 0.062 = x

[HI] = 2x = 0.062 * 2 = 0.124 at equillibrium

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