Some clarifications

[NE2000]

- Basicity constant: will we ever be asked about it? It seems basically (lol) the same as acidity constant but I just want to know whether to try out examples and stuff.

Will add more clarifications required when I get around to it

[Mao]

Probably not, but useful to know.

[nerd]

You don't need to know it specifically, but in an exam they would explain to you what a basicity constant is and expect you apply the knowledge you have about the acidity constant to that situation.

[NE2000]

1. In Nelson Chemistry it says that a K value of around 1000 indicates a reaction goes to completion. In Heinemann it says around 10000. Which is more accurate? And on a similar note, what is the equivalent Ka value (is it just like 1000 x 56 because of [H2O])?

2. Heinemann suggests that all catalysts tend to have the same effect on both the forward and reverse reactions, and hence K is not at all influenced by the presence of a catalyst. But are there catalysts that only promote the forward reaction? Do enzymes do both forward and reverse or are they effective both ways too? I would assume if a catalyst promoted only the forward reaction the K value would be larger.

[nerd]

1. In Nelson Chemistry it says that a K value of around 1000 indicates a reaction goes to completion. In Heinemann it says around 10000. Which is more accurate? And on a similar note, what is the equivalent Ka value (is it just like 1000 x 56 because of [H2O])?

I would be inclined to go with Heinemann on this one.

   If K > 10⁴, the amount of reactants remaining is negligible. Therefore the reaction goes to completion.
   If 10^–4 < K < 10⁴, then there are significant amounts of both products and reactants.
   If K < 10^–4, the amount of products formed is negligible. Therefore the reaction occurs at a negligible extent.

2. Heinemann suggests that all catalysts tend to have the same effect on both the forward and reverse reactions, and hence K is not at all influenced by the presence of a catalyst. But are there catalysts that only promote the forward reaction? Do enzymes do both forward and reverse or are they effective both ways too? I would assume if a catalyst promoted only the forward reaction the K value would be larger.

If a catalyst provides an easier path for the reaction, the path for the reverse reaction is made equally easier. A catalyst will not shift an equilibrium position because both rates are always equally increased. The equilibrium is achieved quicker in time and under easier conditions, however.

[NE2000]

2. Heinemann suggests that all catalysts tend to have the same effect on both the forward and reverse reactions, and hence K is not at all influenced by the presence of a catalyst. But are there catalysts that only promote the forward reaction? Do enzymes do both forward and reverse or are they effective both ways too? I would assume if a catalyst promoted only the forward reaction the K value would be larger.

If a catalyst provides an easier path for the reaction, the path for the reverse reaction is made equally easier. A catalyst will not shift an equilibrium position because both rates are always equally increased. The equilibrium is achieved quicker in time and under easier conditions, however.

What created confusion were enzymes that are specific to say a hydrolysis reaction. They would not also facilitate the condensation reaction (at least I don't think so seeing as their name is often something hydroxylase or something similar). In that case would the equilibrium be affected?

[nerd]

As all catalysts, enzymes do not alter the position of the chemical equilibrium of the reaction. Usually, in the presence of an enzyme, the reaction runs in the same direction as it would without the enzyme, just more quickly. However, in the absence of the enzyme, other possible uncatalyzed, "spontaneous" reactions might lead to different products, because in those conditions this different product is formed faster.

Furthermore, enzymes can couple two or more reactions, so that a thermodynamically favorable reaction can be used to "drive" a thermodynamically unfavorable one. For example, the hydrolysis of ATP is often used to drive other chemical reactions.

Enzymes catalyze the forward and backward reactions equally. They do not alter the equilibrium itself, but only the speed at which it is reached.

For example, carbonic anhydrase catalyzes its reaction in either direction depending on the concentration of its reactants.
(Image removed from quote.) (in tissues; high CO2 concentration)
(Image removed from quote.) (in lungs; low CO2 concentration)
Nevertheless, if the equilibrium is greatly displaced in one direction, that is, in a very exergonic reaction, the reaction is effectively irreversible. Under these conditions the enzyme will, in fact, only catalyze the reaction in the thermodynamically allowed direction.

In other words, enzymes, like catalysts, will generally not effect the position of equilibrium but rather effect the rate of both the forward and reverse reactions equally.

[NE2000]

As all catalysts, enzymes do not alter the position of the chemical equilibrium of the reaction. Usually, in the presence of an enzyme, the reaction runs in the same direction as it would without the enzyme, just more quickly. However, in the absence of the enzyme, other possible uncatalyzed, "spontaneous" reactions might lead to different products, because in those conditions this different product is formed faster.

Furthermore, enzymes can couple two or more reactions, so that a thermodynamically favorable reaction can be used to "drive" a thermodynamically unfavorable one. For example, the hydrolysis of ATP is often used to drive other chemical reactions.

Enzymes catalyze the forward and backward reactions equally. They do not alter the equilibrium itself, but only the speed at which it is reached.

For example, carbonic anhydrase catalyzes its reaction in either direction depending on the concentration of its reactants.
(Image removed from quote.) (in tissues; high CO2 concentration)
(Image removed from quote.) (in lungs; low CO2 concentration)
Nevertheless, if the equilibrium is greatly displaced in one direction, that is, in a very exergonic reaction, the reaction is effectively irreversible. Under these conditions the enzyme will, in fact, only catalyze the reaction in the thermodynamically allowed direction.

In other words, enzymes, like catalysts, will generally not effect the position of equilibrium but rather effect the rate of both the forward and reverse reactions equally.

ok thanks

[NE2000]

3. What's the difference between a coproduct and a by-product?

4. Book's got heaps of general stuff on safety and hazards, the HAZCHEM signs, UN identity numbers, MSDS, lots of laws and stuff. How much should be going onto my notes and how much should be read and then not cared about much?

[NE2000]

Another two:

5. The yellow Fe3+ ion reacts with colourless SCN- ion to form the red [FeSCN]2+ ion. They ask what will be observed when more Fe3+ was added. The answer says it will turn more red (and I understand why it says it will turn more red). But what caused me confusion was that the system will only partially counteract the change. The change is you add more yellow compound, so compared to before won't there still be a net relative increase in concentration of the yellow compound? Am I missing something? Help would be appreciated.

6. When you decrease pressure but everything else remains constant you must be adding an inert gas right? And the reason an inert gas doesn't change equilibrium is because the concentration of substances don't change at all, there's still the same number of mole in the same container? Just confused by one of the answers to a question I was doing where it said that the change in pressure would cause no change in the concentration fraction although the concentration of products would change 

[hyperblade01]

Another one:

5. The yellow Fe3+ ion reacts with colourless SCN- ion to form the red [FeSCN]2+ ion. They ask what will be observed when more Fe3+ was added. The answer says it will turn more red (and I understand why it says it will turn more red). But what caused me confusion was that the system will only partially counteract the change. The change is you add more yellow compound, so compared to before won't there still be a net relative increase in concentration of the yellow compound? Am I missing something? Help would be appreciated.

Well Le Chatelier's Principle in the Heinamann textbook does say:

"If an equilibrium system os subjected to a change, the system will adjust itself to partially oppose the effect of the change"

There would be a net increase of [Fe3+] but as a result, more FeSCN2+ will form and kind of 'balance' the concentrations.

Have you tried drawing those Concentration vs Time graphs? It might help

[TrueTears]

Basically what hyperblade01 said, but think about it, you add more Fe3+ so the system will partially oppose this by favouring the forward reaction, so there will be a increase in the concentration of [FeSCN]2+. However, as the reaction keeps going it will try to reach equilibrium again. It will indeed appear very yellow at the beginning of the reaction (due to the sudden increase, however I assume the question wants what is the colour changing to as it APPROACHES the new equilibrium) so as it REACHES equilibrium it will slowly turn more and more red. [the K stays the same, hence you can consider the relative concentrations of each to be the "same" at the new equilibrium, so the colour will be more red as it reaches equilibrium]

[Collin Li]

Colour is proportional to concentration of the coloured compound and independent of the other concentrations. So ignore the yellow concentration, and focus on the red compound's concentration.

[TrueTears]

A6, by adding a inert gas does not change the VOLUME of the container but it does change the pressure acting on the container.

Think about it like this, PV = nRT

the number of mol of whatever substance stays the same, T is constant, R is constant, so let nRT = k

PV = k



Now,

Now the volume of the container is changed by an external force. By adding an inert gas increases the pressure exerted by it on the sides of the container but there is no external forces acting to make the container shrink or expand. So the total V does not change.

Hence adding a inert gas does nothing, the relative concentrations of the substances in the container stays the same.

With regards to the question you met, yes the pressure could change by adding an inert gas, but this inert gas does not change the total volume of the container hence the relative concentrations stays the same, therefore a change in pressure in this case would do nothing to the concentration fraction.

[dummy]

if a balloon is involved....... ?

With regards to the question you met, yes the pressure could change by adding an inert gas, but this inert gas does not change the total volume of the container hence the relative concentrations stays the same, therefore a change in pressure in this case would do nothing to the concentration fraction.