brightsky's Chem Thread

[lzxnl]

Actually...not a single bit of this stuff is that much beyond VCE level. You know what acidity constants are, you know how to use them with regards to acid-base equilibria and you just need the pKa value of HSO4- of 1.99.
All I've done is use equilibrium constants to numerically find things. Is there anything in my working that is really that complex?

As for the calcium hydroxide stuff, Mao has phrased it much more eloquently than I have; it's an ionic salt, and you learn about the properties of dissolved salts in 1/2 I believe, so that's not really complex either.

If you're referring to the random bit about Al(OH)4 -...well a bit of reading of wiki doesn't hurt from time to time

[brightsky]

Consider the following reaction: Mg(s) + 2HCl(aq) --> MgCl2(aq) + H2(g). Describe three ways in which the rate of this reaction could be monitored.

Any help appreciated!

(So far, I only have "Monitor the pH of the system over time using a pH meter". Is there any way in which either the volume of H2 produced or the disappearance of Mg can be quantitatively measured?)

[SocialRhubarb]

Yep.

You could measure the volume of hydrogen gas produced over time, or measure the decrease in the mass of the system over time in an open system.

[brightsky]

Yep.

You could measure the volume of hydrogen gas produced over time, or measure the decrease in the mass of the system over time in an open system.

Ah I see. Thanks! Although how would you go about making these measurements?

[SocialRhubarb]

Volume of gas could be measured with a gas syringe.

Mass of system can be done if you just conduct your hole experiment in a beaker on a scale.

[brightsky]

Thanks SocialRhubarb!

When deciding how an increase in total volume will influence the position of equilibrium, do we ignore all species which are not gaseous, and focus solely on those that are? For example, consider the reaction: CO2(g) + CaO(s) <--> CaCO3(s). When presenting a  case using LCP, do we ignore CaCO and CaCO3, and reason that since the back reaction produces more gas particles, an increase in total volume will cause the equilibrium position to shift to the left?

[Aurelian]

Thanks SocialRhubarb!

When deciding how an increase in total volume will influence the position of equilibrium, do we ignore all species which are not gaseous, and focus solely on those that are? For example, consider the reaction: CO2(g) + CaO(s) <--> CaCO3(s). When presenting a  case using LCP, do we ignore CaCO and CaCO3, and reason that since the back reaction produces more gas particles, an increase in total volume will cause the equilibrium position to shift to the left?

Yes =)

[brightsky]

In the context of calorimetry, for us to be able to work out the enthalpy change from SHC, mass, and temperature change, is it necessary that the surroundings be made of the one substance? In other words, can we determine the SHC of something that is made of more than one substance?

Thanks!

[Mao]

In the context of calorimetry, for us to be able to work out the enthalpy change from SHC, mass, and temperature change, is it necessary that the surroundings be made of the one substance? In other words, can we determine the SHC of something that is made of more than one substance?

Thanks!

The "calibration factor" is the collection of all the components' . Of course, you could obtain the calibration factor ab initio if you knew the precise mass and of all the different components. In reality, it is far more practical to just measure the overall "effective" heat capacity.

[brightsky]

Ah I see...thanks Mao!

A few other questions:

1) If a question asks us to 'calculate the percentage of the original 0.110 mol of X that has been converted into Y at equilibrium', do we always assume that it is asking for a percentage by mass?
2) Given a concentration-time plot, how do we work out when a catalyst has been added? What signs should we look for?
3) Do values taken from the data booklet affect significance?

[lzxnl]

Ah I see...thanks Mao!

A few other questions:

1) If a question asks us to 'calculate the percentage of the original 0.110 mol of X that has been converted into Y at equilibrium', do we always assume that it is asking for a percentage by mass?
2) Given a concentration-time plot, how do we work out when a catalyst has been added? What signs should we look for?
3) Do values taken from the data booklet affect significance?

Question 1 is...huh? If they want the percentage of the original amount of X that has been converted, you just work out number of moles of X that has reacted / number of moles initially.

Question 2, if both reaction rates suddenly increase by the exact same amount and stay there, a catalyst has been added. It affects both rates equally.

Question 3. If you mean significant figures, I would presume so. Acidity constants, for instance, are rarely given to more than three significant figures; being given to two is generally enough.

[brightsky]

Question 1 is...huh? If they want the percentage of the original amount of X that has been converted, you just work out number of moles of X that has reacted / number of moles initially.

Question 2, if both reaction rates suddenly increase by the exact same amount and stay there, a catalyst has been added. It affects both rates equally.

Question 3. If you mean significant figures, I would presume so. Acidity constants, for instance, are rarely given to more than three significant figures; being given to two is generally enough.

I see, thankyou!


Also, when water is added to a dilute solution of NH3(aq), does the pH of the solution rise and fall?

NH3(aq) + H2O(l) <--> NH4+(aq) + OH-(aq), so if [H2O] increases, then net forward reaction occurs, increasing [OH-] and so raising the pH, no? The solution key suggests otherwise...where have I gone wrong?

Thanks!

[09Ti08]

You're right. However, I think this is more important: you're diluting the solution, which is going to decrease [OH-], this makes it less basic.

[lzxnl]

I see, thankyou!


Also, when water is added to a dilute solution of NH3(aq), does the pH of the solution rise and fall?

NH3(aq) + H2O(l) <--> NH4+(aq) + OH-(aq), so if [H2O] increases, then net forward reaction occurs, increasing [OH-] and so raising the pH, no? The solution key suggests otherwise...where have I gone wrong?

Thanks!

You're right. However, I think this is more important: you're diluting the solution, which is going to decrease [OH-], this makes it less basic.

brightsky, use some common sense here. If you dilute an ammonia solution, there's no way you're going to increase the hydroxide ion concentration. Common sense trumps LCP any day.

NH3(aq) + H2O(l) <=> NH4+(aq) + OH-(aq)

OK, so you add some water. What's the first thing that happens?  The hydroxide ion concentration decreases. Almost instantly. LCP says that a system can only PARTIALLY counteract changes. Here, the dilution has reduced the number of particles per unit volume. All the system can do is partially increase the number of particles per unit volume with a net forward reaction. However, this is only partially opposing. It can't ever recover the change in [OH-] by itself.

Another way of thinking about this:
K=[NH4+][OH-]/[NH3] at equilibrium which is fixed.
Introduce more water. Let's assume you halve the concentrations.
So your new Q value is half of K. The system will try to increase Q to K.
What has changed? The concentration of ammonia has decreased too from the dilution. Due to the mole ratios, concentrations of ammonium ion and hydroxide ion are going to be the same. Now, if [NH3] is lower, then the concentrations of ammonium ion and hydroxide ion at equilibrium cannot possibly be equal to their former values. Previously, [NH3] was higher, so [NH4+] and [OH-] could also be larger. But now that [NH3] is smaller, [NH4+] and [OH-] must be smaller to allow Q to equal K. So despite the fact that the system undergoes a net forward reaction, that just means that after adding the water, your Q is lower than K, so the reaction proceeds forward. The number of MOLES of OH- has increased; the concentration hasn't.

[brightsky]

ahh yes of course, fell right into the trap! thanks guys!