# Chemical equilibrium

1. Aug 12, 2005

there are 2 Questions in which they seem like they should have the same answer, but apparently they dont, which is confusing.

1. 2NH3 <--> N2(g) + 3H2(g) initially, NH3 is added to empty flask. how do rates of forward and reverse reactions change as system proceeds towards eqm?
answer: forward rate decreases and reverse rate increases.

2. 2HBr (g) <--> H2(g) + Br2 (g) initially, HBr is added to empty flask. how do rate of forward reaction and [HBr] change as system proceeds to eqm?
answer: forward rate increases and reverse rate increases.

why are the answers almost opposite? is it because of [HBr] in question2? it seems like the 2 questions are the same..

2. Aug 13, 2005

### GCT

again...what on earth? The question asks for the dynamics of forward reaction and HBR concentration....why does the answer refer to a completely different subject? Have you even posted the corresponding answer? Have you accurately posted the question?

First step in getting help with your question-post the actual question.

3. Aug 13, 2005

### LeonhardEuler

I strongly suspect that this is not the answer to this question. Notice that the question says nothing about the rate of the reverse reaction and asks about [HBr] while the answer references the rate of the reverse reaction but not [HBr]. The answer to the first question can be explained like this: When $NH_3$ is first added, there is no $H_2$ or $N_2$, so there can be no reverse reaction. As the foward reaction proceeds more and more $H_2$ and $N_2$ begin appearing and reacting in the reverse reaction, increasing its rate. At the same time $NH_3$ is being used up, so there is less of it to react, slowing down the rate of the foward reaction.

4. Aug 13, 2005

### Staff: Mentor

These are kinetic approach to equilbrium questions.

First, use Le Chateliers Principle to find out what will happen to the equilibrium in both cases (note volume change during reactions).

Then think how rates of reaction must change in order for the system to move in the direction of new equlibrium state.

And - as it was already said - check out the second question and the second answer, as there is something fishy about them.

5. Aug 14, 2005

The source from which i acquired the question from had a typo. I checked another source with the same question and here is the correct answer: forward reaction decrease as [HBr] decreases. i understand the answer quite clearly now.

but have another query: for reaction kinetics, when monitoring the change in a closed system, would you only look at the reactant side for the change?

also, are homogenous rxns slower than heterogenous?

6. Aug 14, 2005

### Gokul43201

Staff Emeritus
Changes in the reactant side are related to changes in the product side through the stoichiometry of the reaction. For instance :

$$2HBr \leftrightarrow H_2 + Br_2$$

Consuming 2 moles of HBr produces a mole each of H2 and Br2. So, removing any x moles of HBr yields an increase by x/2 moles each of H2 and Br2.

$$\Delta [H_2] = \Delta [Br_2] = -\frac {1}{2} \Delta [HBr]$$

$$\implies \frac {d}{dt} [H_2] = \frac {d}{dt} [Br_2] = -\frac {1}{2} \frac {d}{dt} [HBr]$$

7. Aug 15, 2005

consider: FeO (s) + H2 (g) <--> Fe(s) + H2O (g)

which describes the effect that a decrease in volume would have on the position of equilibrium and the [H2] in the above system?

I was wondering why the [H2] increases even when there is no shift in the equilibrium.

thx

8. Aug 15, 2005

### The Bob

This is because H2 is a smaller molecule than H2O. Equilibrium can shifts with temperature, pressure, increase/decrease of chemcials, volume etc. and also the size or number of molecules.

As the volume has decreases, there is less room so more H2 is produced. However:
This is your question and I cannot fully answer it. To me, there should be a shift to the left due, to the molecule size and an increase in pressure (due to decrease in volume). I am afraid I cannot guess why there is no shift. For this, I apologise.

9. Aug 15, 2005

### Gokul43201

Staff Emeritus
To first order (ie: treating H2O and H2 as ideal gases) a reduction of the volume should have no effect on [H2], because ...

Once again, the provided answer is (partly) wrong. :grumpy:

10. Aug 16, 2005