VCE Chemistry Question Thread

[keltingmeith]

What measure can be used to determine rate of reaction?

Depends. Whichever gives the smallest integer of seconds/minutes/hours/days. If you start to head into months, we generally don't care about it.

[Redoxify]

Sorry that was a misguided question, the question asks how can you determine rate of reaction, sorry!

and also
what affect does adding water have on equilibrium, does le chatliers principle  apply?

[cosine]

I have a few questions:

1. What does the equilibrium constant K, actually indicate? Say K = 50. What does this mean?
2. Why is it that increasing temperature on endothermic reactions will increase K and increase products, but increasing temperature on exothermic reactions decreases the value of K and products?
3. Why does not the addition of catalysts, incresing concentration of reactants and products affect the equilibrium constant?

Many thanks.

[HighTide]

Sorry that was a misguided question, the question asks how can you determine rate of reaction, sorry!

and also
what affect does adding water have on equilibrium, does le chatliers principle  apply?

Add water= dilution. So concentrations of reactants and products will initially decrease. Then the reaction will proceed in whichever direction depending on the number of particles on either sides.
Summary: Add water=dilution= concentrations of all substance initially decrease--> forward or backward reaction becomes more prevalent--> concentration of either products or reactants will increase whilst concentration of the opposite will decrease (depending on which reaction occurs to greater extent)--> Will continue until new equilibrium concentrations established.  This reflects the principle.

I have a few questions:

1. What does the equilibrium constant K, actually indicate? Say K = 50. What does this mean?
2. Why is it that increasing temperature on endothermic reactions will increase K and increase products, but increasing temperature on exothermic reactions decreases the value of K and products?
3. Why does not the addition of catalysts, incresing concentration of reactants and products affect the equilibrium constant?

Many thanks.

Say you have a reaction A+B --> C. . When K equals 1, there is an equal total product concentration and equal reactant concentration.If product concentration>reactant concentration--> K>1  Alternatively, if reactant concentration> product concentration --> K<1
1) So when you have K=50. The numerator must be larger than the denominator. Hence, the total concentration of products must be much larger than the total concentration of reactants. From this, you can deduce that the forward reaction is occurring to a greater extent than the backward reaction as more products are being formed than reactants.

2) In an endothermic reaction: If you increase temperature, more heat energy is present to facilitate the forward reaction. As the energy is required as an input, the forward reaction will be greater as more heat is present. When forward reaction is greater, more products will be formed and K value will be larger than 1.
In an exothermic reaction: Energy is released, therefore it is on the products side of the equation. Thereby, if you increase the temperature, there is more energy to facilitate the backward reaction. When more energy is present, the backward reaction occurs to a greater extent,and more reactants are created than products. Hence, the denominator will be larger, and K will be < 1.

3) Adding a catalyst only changes the reaction rate. If you add a catalyst, the concentration of products and reactants stay the same, the only thing changing is that the reaction will need less time to occur. So at equilibrium, the constant will remain the same as adding a catalyst does not change the position of equilibrium.
Overall, changing concentration of reactants and products cannot change the constant but can alter the position of equilibrium. Using Le Chatelier's principle if you increase the concentration of one, the reaction will alter in a way to bring back dynamic equilibrium. So, decreasing concentration of C, more A and B will be converted to C --> C will increase, A, B, will decrease. Reaction will maintain the same equilibrium constant.

Very sorry for the explanation being verbose but hopefully it helps 
 

[grannysmith]

I have a few questions:

1. What does the equilibrium constant K, actually indicate? Say K = 50. What does this mean?
2. Why is it that increasing temperature on endothermic reactions will increase K and increase products, but increasing temperature on exothermic reactions decreases the value of K and products?
3. Why does not the addition of catalysts, incresing concentration of reactants and products affect the equilibrium constant?

Many thanks.

1. Ratio of [products] to [reactants] at equilibrium.
2. If you increase temp, the system opposes this by favouring the endothermic reaction, because it requires a net input of energy, thus lowering temp. This means the equilibrium position shifts to the products (if we're saying the forward reaction is endothermic), and so [products]:[reactants] increases. Reverse is true for exothermic.
3. Catalysts speed up the reaction both ways, so there's no net direction in which the reaction proceeds. So ratio remains the same. Just remember that catalysts do not affect yield. Only temperature changes K. I don't think going deeper into this concept is necessary.

[cosine]

Add water= dilution. So concentrations of reactants and products will initially decrease. Then the reaction will proceed in whichever direction depending on the number of particles on either sides.
Summary: Add water=dilution= concentrations of all substance initially decrease--> forward or backward reaction becomes more prevalent--> concentration of either products or reactants will increase whilst concentration of the opposite will decrease (depending on which reaction occurs to greater extent)--> Will continue until new equilibrium concentrations established.  This reflects the principle.Say you have a reaction A+B --> C. . When K equals 1, there is an equal total product concentration and equal reactant concentration.If product concentration>reactant concentration--> K>1  Alternatively, if reactant concentration> product concentration --> K<1

1) So when you have K=50. The numerator must be larger than the denominator. Hence, the total concentration of products must be much larger than the total concentration of reactants. From this, you can deduce that the forward reaction is occurring to a greater extent than the backward reaction as more products are being formed than reactants.

2) In an endothermic reaction: If you increase temperature, more heat energy is present to facilitate the forward reaction. As the energy is required as an input, the forward reaction will be greater as more heat is present. When forward reaction is greater, more products will be formed and K value will be larger than 1.
In an exothermic reaction: Energy is released, therefore it is on the products side of the equation. Thereby, if you increase the temperature, there is more energy to facilitate the backward reaction. When more energy is present, the backward reaction occurs to a greater extent,and more reactants are created than products. Hence, the denominator will be larger, and K will be < 1.

3) Adding a catalyst only changes the reaction rate. If you add a catalyst, the concentration of products and reactants stay the same, the only thing changing is that the reaction will need less time to occur. So at equilibrium, the constant will remain the same as adding a catalyst does not change the position of equilibrium.
Overall, changing concentration of reactants and products cannot change the constant but can alter the position of equilibrium. Using Le Chatelier's principle if you increase the concentration of one, the reaction will alter in a way to bring back dynamic equilibrium. So, decreasing concentration of C, more A and B will be converted to C --> C will increase, A, B, will decrease. Reaction will maintain the same equilibrium constant.

Very sorry for the explanation being verbose but hopefully it helps 

1. More products are being formed than reactants. But I thought that equilibrium was the state where the products are being formed at the same rate that the reactants are being formed?

[HighTide]

1. More products are being formed than reactants. But I thought that equilibrium was the state where the products are being formed at the same rate that the reactants are being formed?

Okay so, equilibrium is when the forward and backward reactions occur at an equal rate. So in dynamic equilibrium the concentrations are relatively constant (they don't really change much).. Now say K=50. If I had a solution containing A B and C at constant temperature, concentration of C would be greater than the total concentration of A and B. Volume is constant, so you can imagine this as the moles of C > moles of A and moles of B. You're right. My choice of wording was poor. What I meant to say was, at any time, given that no change is administered, the concentration of product would be greater than the concentration of reactants.
The reaction rates are equal for forward and backward reactions, thereby causing the concentrations to remain relatively stable. The important thing is, in my solution of A,B and C, C would be more prevalent than A and B if the K value for the reaction was 50.

[cosine]

Cheers hightide.

Can someone explain Le Chatelier's law to me please?

[thushan]

It's effectively just a mnemonic that helps you remember what happens to an equilibrium system when you perform a given change on the system.

"If a system originally at equilibrium is subject to a change, the system will act as to partially oppose the change."

Note that this is only true in specific circumstances, namely when you change ONE of the following:

• the concentration of a SINGLE substance; the reaction that is subsequently favoured changes the concentration of that substance in the opposite direction
• the VOLUME of the system; if the volume is increased (and by coincidence pressure decreased), the system will favour the reaction that produces more particles (subsequently, by coincidence, increasing pressure) and if the volume is decreased, the system will favour the reaction that produces fewer particles
• the TEMPERATURE of the system; if you increase the temperature of the system (thereby adding thermal energy into the system), the system will favour the endothermic direction (which subsequently 'sucks out' thermal energy and converts it into chemical energy).

Note that this is merely a memory tool. The reason why equilibrium systems behave that way can be explained by seeing the changes that occur to the reaction quotient (Q) and the equilibrium constant (K) as you perform the change, and as the system subsequently responds to the change.

[cosine]

It's effectively just a mnemonic that helps you remember what happens to an equilibrium system when you perform a given change on the system.

"If a system originally at equilibrium is subject to a change, the system will act as to partially oppose the change."

Note that this is only true in specific circumstances, namely when you change ONE of the following:

• the concentration of a SINGLE substance; the reaction that is subsequently favoured changes the concentration of that substance in the opposite direction
• the VOLUME of the system; if the volume is increased (and by coincidence pressure decreased), the system will favour the reaction that produces more particles (subsequently, by coincidence, increasing pressure) and if the volume is decreased, the system will favour the reaction that produces fewer particles
• the TEMPERATURE of the system; if you increase the temperature of the system (thereby adding thermal energy into the system), the system will favour the endothermic direction (which subsequently 'sucks out' thermal energy and converts it into chemical energy).

Note that this is merely a memory tool. The reason why equilibrium systems behave that way can be explained by seeing the changes that occur to the reaction quotient (Q) and the equilibrium constant (K) as you perform the change, and as the system subsequently responds to the change.

Cheers thushan

So if the equilibrium constant is high, that is more favourable for industrial chemists, as that indicates that the concentration of the products is greater than the concentration of the reactants at any given time?

[keltingmeith]

Sorry that was a misguided question, the question asks how can you determine rate of reaction, sorry!

Start the reaction and a stopwatch. Wait until the reaction has finished. Stop the stopwatch. You now know how long it takes for the reaction to complete (that is, the rate of reaction).

Of course, in VCE, we more care about how we can increase the rate of reaction, as opposed to calculating or measuring it.

[cosine]

Cheers thushan

So if the equilibrium constant is high, that is more favourable for industrial chemists, as that indicates that the concentration of the products is greater than the concentration of the reactants at any given time?

anyone?

[Splash-Tackle-Flail]

anyone?

Yes, if the product is the desired substance, than a higher equilibrium constant would indicate a greater yield of the product from the reaction, and more money for the industrial chemists.

But then they have to deal with other factors as well like cost of producing the reaction, and the rate of the reaction (e.g. in an exothermic reaction, to increase K, scientists could decrease the temperature, but then that'll slow down the reaction rate so they got to compromise)

[cosine]

Few questions about equilibria:

1. Predict the effect of the changes on the position of each equilibrium: Removal of from the equilibrium:

2. For the exothermic reaction: , how would you alter the temperature of the equilibrium in order to product a net forward reaction? I understand that because the forward reaction is exothermic, but what has this to do with decreasing the temperature to produce the forward reaction?

3. An equilibrium mixture containing: The volume of the container was increased at constant temperature and a new equilibrium was established. How would the concentration of be altered? I would say that because the pressure is being decreased, the more particles (moles) is favourable, and hence the forward reaction would increase and so the production of would increase. That being said, although the particles of nitrogen dioxide are being relatively increased, the volume is also increasing so does this actually mean that the concentration is increasing ?

Cheers.

[warya]

I'm confused about changing the equilibrium position. My text says that by adding reactant (or imposing any other change excluding temperature) a new equilibrium is established. Does this mean there is a new value of K?