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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 \rm HI(g) is represented by the equation

\rm 2HI(g)\rightleftharpoons H_2(g)+I_2(g)

The following experiment was devised to determine the equilibrium constant of the reaction.

\rm HI(g) is introduced into five identical 400-\rm cm^3 glass bulbs, and the five bulbs are maintained at 623 \rm K.The amount of \rm I_2 produced over time is measured by opening each bulb and titrating the contents with 0.0150 M \rm Na_2S_2O_3(aq). The reaction of \rm I_2 with the titrant is

 

What is the value of K_c for the decomposition of \rm HI at 623 \rm K???? 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

 

 


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