# Homework Help: Equilibrium problem - right method, wrong answer

1. Jul 18, 2009

### ghostanime2001

1. The problem statement, all variables and given/known data
Keq = 100 at a certain temperature for the reaction

CH4(g) + 2 H2S(g) $$\rightleftharpoons$$ CS2(g) + 4 H2(g)

Some CH4 and H2S were introduced into a 1.0 L bulb and at equilibrium 0.10 mol of CH4 and 0.30 mol of H2S were found. What was [CS2] at equilibrium?
2. Relevant equations

$$K_{eq} = \frac{[CS_{2(g)}][H_{2(g)}]^{4}}{[CH_{4(g)}][H_{2}S_{(g)}]^{2}}$$

3. The attempt at a solution
I know that I'm not given any information about starting or changing amounts so therefore an ICE table would not work. I think to calculate the [CS2]:

$$100 = \frac{^{4}}{[0.1][0.3]^{2}}$$

$$100 = \frac{s^{5}}{0.009}$$

$$s^{5} = 0.9$$

$$s = 0.9791$$

But, the answer from the worksheet is 0.323 M. What am I doing wrong. I've arrived at the answer using the right method but the answer is wrong!! can you please help point out my mistake. Thanks

2. Jul 18, 2009

### Ygggdrasil

for every mole of CS2 produced, how many moles of hydrogen gas must be produced?

3. Jul 18, 2009

### ghostanime2001

I don't understand. CS2 reacts with hydrogen gas. It doesn't produce hydrogen gas.

4. Jul 18, 2009

### ghostanime2001

4 moles of hydrogen gas is produced for every mol of CS2 produced

5. Jul 18, 2009

### ghostanime2001

btw, do i have to find the limiting reactant is this question?

6. Jul 18, 2009

### Ygggdrasil

Correct. But, your expressions for [CS2] and [H2] do not reflect this fact. (e.g. if [CS2] = s, then [H2] is not also s)

7. Jul 19, 2009

### ghostanime2001

4s then?

8. Jul 19, 2009

### Ygggdrasil

Yes.

9. Jul 21, 2009

### ghostanime2001

A reaction mixture is at equilibrium according to the equation: LaCl3(s) + H2O(g) $$\rightleftharpoons$$ LaClO(s) + 2HCl(g). Some more HCl(g) is then added such that when equilibrium is re-established the amount of HCl(g) in the mixture is found to have doubled. By what factor will the amount of H2O in the system have been multiplied?

$$\left[ \mbox{Hint: Write an expression for Keq in terms of unknown concentrations, before and after adding HCl. Then: K_{eq}\mbox{(before)} = K_{eq}\mbox{(after)}} \right]$$

$$\begin{tabular}{|c|c|c|c|c|} \hline & LaCl_{3} & H_{2}O & LaClO & 2HCl \\ \hline S & & & & Y \\ \hline \Delta & -2X & -2X & +2X & +2X \\ \hline E & & & +2X & Y+2X \\ \hline \end{tabular}$$

10. Jul 22, 2009

### Ygggdrasil

Start by writing out the expression for Keq in terms of the concentrations of each reactant and product.

11. Jul 22, 2009

### ghostanime2001

$$K_{eq} = \frac{[HCl]^{2}}{[H_{2}O]}$$

I am not given any information of starting, changing amounts. I could have X moles of both LaCl3 and H2O or neither and have moles of products only OR I can have concentrations of one only. I do not know which. I can also have either products increasing concentration or reactants increasing concentrations. I do not know any information of the "how" of initial amounts will progress as time goes on. There must be some other way of doing this problem. Some way which i am misinterpreting. Please help. Let's take this step by step.

Last edited: Jul 22, 2009
12. Jul 22, 2009

### Ygggdrasil

You have all the information you need to solve the problem. Follow the hint and write an expression for Keq before and after the reaction.

13. Jul 22, 2009

### ghostanime2001

$$K_{eq(before)} = \frac{[2x]^2}{x}$$
$$K_{eq(after)} = \frac{[2(2x +s)]^2}{4x} = \frac{16x+4s}{4x)$$

?????? I Really do no understand.

14. Jul 22, 2009

### Ygggdrasil

Let's start with a very basic equation before we start plugging in variables and making things complicated:
$$\frac{[HCl]_i^2}{[H_2O]_i} = \frac{[HCl]_f^2}{[H_2O]_f}$$

Now let's consider what the problem tells us about these values and what relationships the problem asks us to derive. The problem statement says that [HCl]f = 2[HCl]i and asks you to derive a relationship between [H2O]i and [H2O]f.

15. Jul 22, 2009

### ghostanime2001

okay....

$$\frac{[HCl]_i^2}{[H_2O]_i} = \frac{[4HCl]_i^2}{[H_2O]_f}$$

now what ?

Last edited: Jul 22, 2009
16. Jul 22, 2009

### Ygggdrasil

....

17. Jul 22, 2009

### ghostanime2001

errr..... ????

18. Jul 22, 2009

### Ygggdrasil

Let's say [HCl]i = x and [H2O]i = y. Solve for [H2O]f in terms of x and y (you should see that x will cancel).

19. Jul 22, 2009

### ghostanime2001

Why not just create an ICE table?

20. Jul 22, 2009

### Ygggdrasil

You could create an ICE table, but because none of the initial concentrations are known, it will not be so helpful. This question is less of a question about stoichiometry and using ICE tables and more of a question testing basic understanding of the equilibrium constant.

21. Jul 23, 2009

### ghostanime2001

The only thing i know about the equilibrium constant is that its a ratio of concentration of products over reactants. An equilibrium constant more than 1 means more products, less reactants and an equilibrium constant less than one means less products and more reactants. I also know what direction equilibrium will shift when upset given Q and K. What else could I not know about K that`s giving me trouble here.

22. Jul 23, 2009

### ghostanime2001

I like to solve for the unknown using 1 variable x rather than x and y like you showed, Ygggdrasil.
2 variables is confusing for me in this question because i have to look back and forth to see what x and y are. Even though what you say here
is true even though i didn't take into account of making
$$\frac{[HCl]_i^2}{[H_2O]_i} = \frac{[HCl]_f^2}{[H_2O]_f}$$

$$\frac{x^2}{[H_2O]_i} = \frac{(2x)^2}{[H_2O]_f}$$ <== is "2x" really make sense in this case even though this quantity can either +2x or -2x of it which depends what concentration of reactants i have or not ??

$$[H_2O]_f x^2 = 4x^2 [H2O]_i$$

$$[H_2O]_f = \frac{4x^2}{x^2 } [H2O]_i$$

$$[H_2O]_f = 4[H2O]_i$$

Correct or no?

23. Jul 24, 2009

### Ygggdrasil

According to the wording of the problem, you would add HCl and let the system return to equilibrium until [HCl]f = 2[HCl]i. When you add more HCl to the system, some of the HCl will react with the LaClO to produce reactants, while some of the HCl will remain unreacted so that equilibrium is reestablished. This would decrease the amount of LaClO and increase the amount of H2O and LaCl3. The only problem would be if there is not sufficient LaClO, but for the purpose of this question, you can probably assume there is enough.

Correct.

24. Aug 4, 2009

### ghostanime2001

Just trying to be as accurate as i can.