Areaction mixture in a 3.67 l flask at a certain temperature initially contains a mixture of 0.763 grams of h2 and 96.9 grams of i2. at equilibrium, the flask contains 90.4 grams if hi. calculate the equilibrium constant for this reaction at this temperature.
show your work using an ice table.

H2 + I2 = 2HI

we express Keq as follows:

Keq = [HI]^2 / [H2] [I2]

Where the terms represents the concentrations of the substances involved. THe concentrations are as follows:

H2 = 0.763 ( 1 / 18.02) / 3.67 = 0.0115 M

I2 = 96.9 ( 1 / 253.8) / 3.67 = 0.1040 M

HI (at equilibrium) = 90.4 ( 1 / 127.91) / 3.67 = 0.1956 M 

By the ICE table, we can calculate the equilibrium concentrations,
        H2                       I2                           HI
I      0.0115              0.1040                        0
C      -x                        -x                         +2x
 
E  0.0115-0.0963   0.1040-0.0963       0.1926 ===> x = 0.0963

Keq = (0.1926)^2 / (0.0077) (0.0848)
Keq = 56.81

a is the answer to this question

explanation: