Chemistry 3/4 2013 Thread

[Stick]

Oh good. I wasn't really sure about how the unit would change and assumed that its power would change. I'm glad that I'll only have to evaluate the concentration fraction. Thanks.

[Nato]

quick solubility question

in this question we are required to find the mass of solute to make up the following solution

of

thanks

[psyxwar]

quick solubility question

in this question we are required to find the mass of solute to make up the following solution

of

thanks

0.25M solute means 0.25 moles per litre.

1.5L * 0.25M = 0.375 mole

m=nM
m= 0.375*M()
m= 0.375(2+32+16*4) = 0.375*98
m = 36.75g

[Jeggz]

quick solubility question

in this question we are required to find the mass of solute to make up the following solution

of

thanks

We know







grams

EDIT : Beaten

[Edward21]

Question how does the pH of water rise as temperature rises, at 25C the pH is 7, as the reaction is endothermic, when you heat it up there is a net forward reactions, ok I get that much, so the textbook describes how pH lowers, but then in its questions it says the pH is neutral and there is no pH change as the OH- will equal the H3O+...!? I'm so confused 

[scribble]

pH=-log[H+]
if the reaction H2O <=> H+ + OH- goes forwards, [H+] and [OH-] both increase.
since theyre in the same mol ratio, [H+]=[OH-] which means that the the water is neutral.
but since pH=-log[H+], and [H+] has increased, that means that the pH decreases.

so pure water is ALWAYS neutral, but will have different pHs depending on the temperature. at 25'C, its pH is 7.

[Edward21]

pH=-log[H+]
if the reaction H2O <=> H+ + OH- goes forwards, [H+] and [OH-] both increase.
since theyre in the same mol ratio, [H+]=[OH-] which means that the the water is neutral.
but since pH=-log[H+], and [H+] has increased, that means that the pH decreases.

so pure water is ALWAYS neutral, but will have different pHs depending on the temperature. at 25'C, its pH is 7.

OK. That was confusing but I understand. By definition if OH equals H+ then it's neutral, however because technically speaking the H+ does increase, that alters the pH is lower as that increase. Right, I got it. THANKYOU 

[Edward21]

If anyone has checkpoints 2011 then I'm referring to questions 2 and 14.
Q2) Nitrogen and Hydrogen gas forming ammonia, exothermic. N2 + 3H2 <-> 2NH3 Which would not cause the rate of the forward reaction to increase? A) Increasing the temperature B) Increasing the pressure C) Adding a suitable catalyst D) Adding an inert gas. Answer is D.

Q14) CO + 2H2 <-> CH3OH, exothermic using catalyst. Which would give a maximum yield of methanol?
A) Increase the temperature B) Lower the pressure C) Use a more effective catalyst D) Condense methanol and recycle gases. Answer is D.

For Q2) Wouldn't increasing the temperature cause a net backward reaction to cool the system down? How would that cause the rate of the forward reaction to increase???

Q14) I totally understand this one, even though it's very similar to the previous one. Here increasing temperature would reduce the yield of methanol as the system wants to partially oppose this change by cooling it down. How come the same logic isn't applicable to Q2? 

[Alwin]

If anyone has checkpoints 2011 then I'm referring to questions 2 and 14.
Q2) Nitrogen and Hydrogen gas forming ammonia, exothermic. N2 + 3H2 <-> 2NH3 Which would not cause the rate of the forward reaction to increase? A) Increasing the temperature B) Increasing the pressure C) Adding a suitable catalyst D) Adding an inert gas. Answer is D.

Q14) CO + 2H2 <-> CH3OH, exothermic using catalyst. Which would give a maximum yield of methanol?
A) Increase the temperature B) Lower the pressure C) Use a more effective catalyst D) Condense methanol and recycle gases. Answer is D.

For Q2) Wouldn't increasing the temperature cause a net backward reaction to cool the system down? How would that cause the rate of the forward reaction to increase???

Q14) I totally understand this one, even though it's very similar to the previous one. Here increasing temperature would reduce the yield of methanol as the system wants to partially oppose this change by cooling it down. How come the same logic isn't applicable to Q2? 

Q2 Increasing the temperature means that both forward and reverse reactions increase (higher temperature, greater KE etc etc) So even though the net reaction is backwards, the forward reaction increased too.

Q14 This is asking for the NET increase, where as Q2 was asking for just the forward reaction.

If you're still confused, post again and ill look around for a pic to explain it for you

[Edward21]

Q2 Increasing the temperature means that both forward and reverse reactions increase (higher temperature, greater KE etc etc) So even though the net reaction is backwards, the forward reaction increased too.

Q14 This is asking for the NET increase, where as Q2 was asking for just the forward reaction.

If you're still confused, post again and ill look around for a pic to explain it for you

Oh ok, so even if the temperature rises, the direction of the reaction which the equilibrium not doesn't favour, is still going to proceed at a faster rate, despite the position of the eq. moving opposite it?

[lzxnl]

Question how does the pH of water rise as temperature rises, at 25C the pH is 7, as the reaction is endothermic, when you heat it up there is a net forward reactions, ok I get that much, so the textbook describes how pH lowers, but then in its questions it says the pH is neutral and there is no pH change as the OH- will equal the H3O+...!? I'm so confused 

Are you talking pure water? There's an equation called the Van't Hoff equation which VCAA unfortunately don't teach.
It goes like this.

Now from memory the enthalpy change for the reaction of H+ and OH- is -55 kJ/mol
So plug in and your value for into the equation. Now you have a relationship between and temperature. As the water remains neutral, square root the equilibrium constant to work out the numerical value of concentration in M of H+. Take log and you're done.

If anyone has checkpoints 2011 then I'm referring to questions 2 and 14.
Q2) Nitrogen and Hydrogen gas forming ammonia, exothermic. N2 + 3H2 <-> 2NH3 Which would not cause the rate of the forward reaction to increase? A) Increasing the temperature B) Increasing the pressure C) Adding a suitable catalyst D) Adding an inert gas. Answer is D.

Q14) CO + 2H2 <-> CH3OH, exothermic using catalyst. Which would give a maximum yield of methanol?
A) Increase the temperature B) Lower the pressure C) Use a more effective catalyst D) Condense methanol and recycle gases. Answer is D.

For Q2) Wouldn't increasing the temperature cause a net backward reaction to cool the system down? How would that cause the rate of the forward reaction to increase???

Q14) I totally understand this one, even though it's very similar to the previous one. Here increasing temperature would reduce the yield of methanol as the system wants to partially oppose this change by cooling it down. How come the same logic isn't applicable to Q2? 

May have been said before, but increasing temperature always increases reaction rate (don't look up negative activation energy on Wiki...please). Therefore increasing the temperature increases the forward rate, just increases the back rate even more. The proof is slightly messy.
Don't get confused about equilibrium shifts and reaction rates. Equilibrium shifts mean that one reaction increases MORE than the other one, but it says nothing about whether both of them increase at the same time. Here's an example. If you and I are pulling on a piece of rope and we're both initially pulling with 100 N, the rope isn't going to move, i.e. equilibrium. Doesn't mean there's no force, just no net force. However, if you then pull with 300 N while I only pull with 200 N, the rope will move towards you. Equilibrium has been broken in your favour, but it doesn't mean that I wasn't pulling harder. Now replace forces with reaction rate and pulling the rope with chemical reactions.

[Edward21]

Are you talking pure water? There's an equation called the Van't Hoff equation which VCAA unfortunately don't teach.
It goes like this.

Now from memory the enthalpy change for the reaction of H+ and OH- is -55 kJ/mol
So plug in and your value for into the equation. Now you have a relationship between and temperature. As the water remains neutral, square root the equilibrium constant to work out the numerical value of concentration in M of H+. Take log and you're done.


May have been said before, but increasing temperature always increases reaction rate (don't look up negative activation energy on Wiki...please). Therefore increasing the temperature increases the forward rate, just increases the back rate even more. The proof is slightly messy.
Don't get confused about equilibrium shifts and reaction rates. Equilibrium shifts mean that one reaction increases MORE than the other one, but it says nothing about whether both of them increase at the same time. Here's an example. If you and I are pulling on a piece of rope and we're both initially pulling with 100 N, the rope isn't going to move, i.e. equilibrium. Doesn't mean there's no force, just no net force. However, if you then pull with 300 N while I only pull with 200 N, the rope will move towards you. Equilibrium has been broken in your favour, but it doesn't mean that I wasn't pulling harder. Now replace forces with reaction rate and pulling the rope with chemical reactions.

Sort of like what I said before? Where for an exothermic reaction, if the temperature increases, even though both rates increase (so the forward would actually increase due to a higher average proportion of particles with adequate kinetic energy to overcome the Ea barrier and whatnot) if the pos. of eq. shits to the left, that would indicate that even though both rates are going faster at a higher temperature, the backwards reaction is moving at a faster rate to cool off the closed system? Basically, high temp, both rates faster, however due to the eq. position one of the rates is even fast than the other as the system tries to partially oppose the change? Is this right, ahhhh! 

[scribble]

^you're correct, yes.

[lzxnl]

Sort of like what I said before? Where for an exothermic reaction, if the temperature increases, even though both rates increase (so the forward would actually increase due to a higher average proportion of particles with adequate kinetic energy to overcome the Ea barrier and whatnot) if the pos. of eq. shits to the left, that would indicate that even though both rates are going faster at a higher temperature, the backwards reaction is moving at a faster rate to cool off the closed system? Basically, high temp, both rates faster, however due to the eq. position one of the rates is even fast than the other as the system tries to partially oppose the change? Is this right, ahhhh! 

I wouldn't say "due to the equilibrium position" as the shift in equilibrium position is the effect, not the cause. The cause is the fact that identical increases in temperature affect exothermic reaction rates less than that for endothermic reaction rates.
I've mentioned this before. Reaction rate is proportional to . If you differentiate this with respect to temperature, you get . As endothermic reactions have higher activation energies than exothermic reactions (draw out the reaction energy profile and you'll see), the derivative is larger. Therefore this exponential term increases more for endothermic reactions than exothermic reactions for a fixed increase in T. Similarly, for a fixed decrease in T, the exponential term decreases more for endothermic reactions. THIS is what causes the "equilibrium position" shift.

Le Chatelier works, but IMO it doesn't do a great job of explaining why on earth the system tries to offset the disturbance. The mathematics does.

[Edward21]

I wouldn't say "due to the equilibrium position" as the shift in equilibrium position is the effect, not the cause. The cause is the fact that identical increases in temperature affect exothermic reaction rates less than that for endothermic reaction rates.
I've mentioned this before. Reaction rate is proportional to . If you differentiate this with respect to temperature, you get . As endothermic reactions have higher activation energies than exothermic reactions (draw out the reaction energy profile and you'll see), the derivative is larger. Therefore this exponential term increases more for endothermic reactions than exothermic reactions for a fixed increase in T. Similarly, for a fixed decrease in T, the exponential term decreases more for endothermic reactions. THIS is what causes the "equilibrium position" shift.

Le Chatelier works, but IMO it doesn't do a great job of explaining why on earth the system tries to offset the disturbance. The mathematics does.

Oh yes I see, your explanation is so methods haha! But I understand, it seems there's few gaps in the course where we're fed knowledge to rote learn from the textbook rather than to understand the further concepts behind them.. the limits of the VCE curriculum?