A reaction mixture in a 3.51 l flask at a certain temperature initially contains a mixture of 0.864 grams of h2 and 103.7 grams of i2. at equilibrium, the flask contains 92.3 grams if hi. calculate the equilibrium constant for this reaction at this temperature. show your work.

ExYou must use the I.C.E. table for this problem. First, you must convert the grams given into M (molarity)which is in(moles/L).

Molar Masses:

H2= 1.01x2 = 2.02g/mol

I2= 126.90x2 = 253.80g/mol

HI= 1.01+126.90=127.91g/mol

Converting to Moles:

0.864g of H2/(mol/2.02g)=0.432

103.7g of I2/(mol/253.80g)=0.409mol

92.3g of HI/(mol/127.91g/mol)=0.722mol

Converting to M:

H2=0.432mol/3.51L=0.123M

I2=0.409mol/3.51L=0.117M

HI=0.722mol/3.51L=0.206M

You must have M (at least for this question) to make an I.C.E. table.

I.C.E. Table:(M)

H2 I2 2HI Kc

I:0.123 0.117 0.00 0.00

C:-0.0965 -0.0965 +0.193 (an increase)

E:0.0065 0.0075 0.206 ?

Solve for Equilibrium Constant (Kc):

(0.206)^2/(0.0065)(0.0075)

Kc=475

please check calculation