# Chemical equilibrium and Le Chatelier's principle

$$\ce{energy + 2KClO3 (s) <=> 2KCl(s) + 3O2 (g)}$$

The reaction will be more active in the reverse direction if:

1. $\ce{O_2}$ is added
2. more $\ce{KCl}$ salt is added.
3. More $\ce{KClO3}$ is added
4. Increasing the temperature.

My thoughts:

I think (1) and (2) are both right. But the book states that only (1) works and I can't find reason for that searching google was useless.

$K_c = [\ce{O2}]^3$, so if we increased $\ce{KCl}$ or decreased $\ce{KClO3}$ it won't affect the equilibrium?

$$\ce{Fe(OH)(s) <=> Fe^{2+} + 2OH^-}$$ The reaction will be more active in the reverse direction if : ... is added

1. $\ce{KOH}$
2. $\ce{Na_2S}$
3. $\ce{Fe(OH)_2}$
4. $\ce{Fe(NO3)2}$

Won't it be (1) and (4)? The book says it's (4) only.

• Why do you think (1) and (2) are correct?
– user15489
May 26, 2015 at 19:48
• Can you write out the formula for the equilibrium constant? May 26, 2015 at 19:49
• Because if I added KCL le chattele ( never written his name right) says the reaction will go in the direction which makes it get back to equilibrium and so will happen if we added o2. maybe it's misunderstanding from me.
– user16485
May 26, 2015 at 19:52
• Ok , Kc = [0_2]^3
– user16485
May 26, 2015 at 19:53
• Bro, I think I am right check this : chemguide.co.uk/physical/equilibria/kc.html (Writing an expression for Kc for a heterogeneous equilibrium) You will see "The important difference this time is that you don't include any term for a solid in the equilibrium expression." this too bouman.chem.georgetown.edu/S02/lect8/lect8.htm
– user16485
May 26, 2015 at 20:13

1) Increasing Oxygen pressure will increase the reverse rate because to reestablish equilibrium some of the oxygen should be consumed.
2) Out of system due to precipitation.
3) Out of system due to precipitation.
4) This is endothermic reaction. So increasing temperature will increase the forward rate to consume the added heat.

• "Oxygen pressure" ??
– user16485
May 26, 2015 at 20:20
• Adding gas phase oxygen means increasing the partial pressure of oxygen. May 26, 2015 at 20:21
• Because we are increasing the number of atoms in Unit of volume ? p(ro) = m/Vol
– user16485
May 26, 2015 at 20:23
• Yeah. Think about it from mathematical point of view: You have a constant K which has numerical value of 100. K is also equal to ${O_2^3}$. So that means ${O_2}$ is 4.64 bar roughly. Now you increase it to 5 bar and the equilibrium wants it to stay at 4.64 bar. So the equilibrium will consume 0.35 bar worth of oxygen molecule to reestablish it to 100. May 26, 2015 at 20:28
• A reference that helps explain what is happening with the oxygen is at chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/…
– user15489
May 26, 2015 at 20:28