calculating the equilibrium concentrations

Question

A mixture of 0.460 M H2O, 0.393 M Cl2O, and 0.683 M HClO are enclosed in a vessel at 25 °C.

Calculate the equilibrium concentrations of each gas at 25 °C.

H2O (g) +Cl2O(g) --><--2HOCl(g)

for H2O Cl2O and 2HOCl

kc= 0.090 at 25 degrees C

Answer 1

1. Make an ICE table . (Initial, Change, Equilibrium)
2. You know the INITIAL concentrations. Those are given in the first line of the problem.
3. You don't know the CHANGE, so that's an x for every mole. Remember to account for stoichiometry in the change!
4. the EQUILIBRIUM concentrations will be expressions in terms of x: Initial concentration MINUS the stoichiometrically correct number of x's (Or PLUS, if the substance is being formed. Remember: if Reactants go away, products must be formed and vice versa. )
5. Write an expression for the equilibrium constant and plug your equilibrium values into that and solve for x.
6. then finally use the value of x to figure out the equilbrium concentrations using the expressions you wrote in part 1.c.

Okay so those are the steps, now I'll show you the numbers.

1. ICE Table:

H2O + Cl2O ---> 2HOCl

Initial: 0.460 0.393 ---> 0.683

Change: +x +x -2x

Equilibrium: 0.460+x 0.393+x 0.683-2x

Notice the stoichiometric correctness of the x's

All concentrations are in M

I don't actually know beforehand which way this reaction is going to shift. I used Q (the value of K before the reaction) to figure this out. In this case, the value is 2.5, which is much greater than the value of Keq (0.09). This means that the reaction will shift left (products are consumed and reactants are produced), and I wrote the Change that way in the ICE table.

IF you don't bother calculating Q, and just arbitrarily pick a direction ,and that direction turns out to be incorrect - there is no need to worry, because the math will correct you in the end. Try it and prove that to yourself sometime!.

2. Kc = [Products]/[Reactants]

THe expression for Kc for this reaction is :

[HOCl]^2 / [H2O][Cl2O]

In this expression H2O is included because it is a gas and not in its standard state (i.e., liquid).

Kc =[HOCl]^2 / [H2O][Cl2O]

0.090 = [0.683-2x]^2 / ([0.460+x] [0.393+x])

When I simplify this hideous equation I get the quadratic:

3.91x^2 - 2.809 x + 0.4502 = 0

I plugged those into a quadratic equation solver and got two values for x:

0.477 and 0.241

3. The third step is to plug those values for x into the expressions for equilibrium concentration that we wrote in 1.c.

It quickly becomes apparent that 0.477 is NOT The correct value for x, because that would give a NEGATIVE equilibrium concentration for HOCl. (0.683 - 2*0.477 = -.271 that is no good).

SO x = 0.241 and the equilibrium concentrations are:

H2O = 0.460 + .241 = 0.701

Cl2O = 0.393 + .241 = .634

HOCl = 0.683 - 2*.241 = .201

Check your work by plugging those values into K and seeing if you get the correct answer (0.09):

Kc =[HOCl]^2 / [H2O][Cl2O] = (.201 ^ 2) / (.701*.634) = 0.0909 which is 0.09, in terms of sig figs! Hurray!