# solving for Kc of the decomposition of HI at 623K

Submitted by Alyssa.the.chemme on Sat, 03/23/2013 - 11:43

The decomposition of is represented by the equation The following experiment was devised to determine the equilibrium constant of the reaction. is introduced into five identical 400- glass bulbs, and the five bulbs are maintained at 623 .The amount of produced over time is measured by opening each bulb and titrating the contents with 0.0150  . The reaction of with the titrant is

What is the value of for the decomposition of at 623 ???? This is the question I am trying to solce for.

I know that Q=([H2]*[I2])/[HI]^2

Concentrations---   H2= 5.378*10^-4 the concentation of I2=H2

Concentration of HI= 4.398*10^-3

Experimental data

 Bulb Initial mass of ( ) Time( ) Volume of titrant( ) 1 0.300 2 20.96 2 0.320 4 27.90 3 0.315 12 32.31 4 0.406 20 41.50 5 0.280 40 28.68

Since the system is at equilibrium, Kc will equal Q.  So, since you now have your concentrations (which was the hard part), plug those into the equation:

Q=[Products]^coefficient/[Reactants]^cofficient=Kc

In this case the equation will be:

Q=Kc= [H2][I2]/[HI]^2  *Note that since HI had a coefficient of 2 in the reaction, you raise [HI] to the second power

Now, plug in your values that you solved for:

Q=Kc=[5.378*10^-4][5.378*10^-4]/[4.398*10^-3]2

Q=Kc= 1.5*10^-2

Hope that helps.  Make sure you understand how to work through this and why Kc=Q in this scenario--because it is in equilibrium.

K

c =1.50×10−2

How did you find the concentrations for HI, H2, and I2?

The amount of I2 can be determined from the volume and molarity of the Na2S2O3 using the equation

2   S2O3-2   +  I2    --->  2 I-   +   S2O4 2-

A starch solution indicator is used which turns black/purple in the presence of the I2, once all the I2 is used up, the solution turns a slight yellow color.

So using the volume and molarity of the S2O3 2- used, you can calculate the moles of S2O3 2- and then the moles of I2 present at equilibrium - .5 x moles of S2O3 2-  (a).

Once that is done you can set up an ICE chart:

HI               H2              I2

Initial                                 *                 0                0     (convert moles to g)

Change                            -2a                +a            +a     (since 1 mole of H2 must be produced for every mole of I2)

Equilibrium                       * - 2a             +a           +a