help with questions

[Rosie]

Q. Consider the reaction: 2A_(s) + 3B_(l) C_(aq) + 4D_(aq)
Which of the following is the correct expression for the equilibrium constant?

A.

B.

C.

D.


I would have thought that D is the only correct answer but it states that B or D is the correct answer. How is that possible?

[shinny]

I assume its got something to do with the states of A and B since they're different from that of C and D. However, as to why they're correct...erm....someone else explain? I'm not too great on technicalities in chem <.<

[Collin Li]

Liquids and solids have a nearly fixed density. For example, you can go and look up the density of water and ice. Since densities have the unit of "mass per volume," concentrations have the unit "amount per volume," and "mass per amount" is constant (the molar mass), then we know density is pretty much just a measure of concentration. So the concentration of liquids and solids are not changing much.

On the other hand, you can't look up the density of steam (gas). It varies depending on the variables in the ideal gas equation. Also, aqueous solutions have largely variable concentrations too. I can dissolve either a 1000 ions into enough water, or just 1 ion, and the amount will change, but the volume will not change much - as a result the concentration can vary vastly.

With liquids and solids, if I measure the volume occupied by 1 molecule compared to 1000 molecules, the volume will be 1000 times as large in the latter case, which means overall, the concentration has not changed.

Hence, the answer B is a good approximation, because the solid species, A, and the liquid species, B, have relatively fixed concentrations, and only the species C and D will change enough to affect equilibrium.

[Pandemonium]

Ah, thank you Coblin.
The TSFX lecturer said ignore solids and I seriously had no idea why she said that.

[Rosie]

q. The best description of the effect of a catalyst on a chemical reaction is that it:

A. lowers the activation energy of the forward reaction without changing the Ea of the reverse reaction
B. lowers the Ea of the forward reaction and raises the activation energy of teh reverse reaction
C. lowers the Ea of both forward and reverse reactions by the same amount
D. lowers the Ea of the reverse reaction without changing the Ea of the forward reaction.

would everyone agree that the answer is C, however it says that the answer is A. How is this possible? 


Another question:
Q. considering the reaction: Hg_(l) + 0.5I_2(g) ----> HgI_(g) 

i.) what is the Ea for the reaction 0.5Hg_(l) + 0.25I_2(g) ----> 0.5HgI_(g)?
ii.) what is the Ea value for the reaction 2HgI_(g)  ------> 2Hg_(l) + I_2(g)?

thanks

[bec]

I got C for the first question too...

Was there a graph for the second question?

[onlyfknhuman]

Q1. i would think it is C too

Q2. Yeah, is there some kind of graph? , for enthalpy for the equations?

[dusty_girl1144]

Q1. i would think it is C too

Q2. Yeah, is there some kind of graph? , for enthalpy for the equations?

q1. a catalyst "runs out" eventually yea? and it lowers activation energy for forward and backwards reaction right?

q2. either way ud need delta H for this?
otherwise u could just delta H halves for i and is doubled and reciprical it for the second? :s... i wouldnt have a clue if its right?

[Collin Li]

a catalyst "runs out" eventually yea?

In theory, no - it is not consumed at all during the reaction. In practice, catalysts can become polluted over time (via side reactions that do involve the consumption of the catalyst, for example), and lose their ability to catalyse.

[onlyfknhuman]

a catalyst "runs out" eventually yea?

In theory, no - it is not consumed at all during the reaction. In practice, catalysts can become polluted over time (via side reactions that do involve the consumption of the catalyst, for example), and lose their ability to catalyse.

YOUR KIDDING >_<. All the questions that involve catalysts, have stated that it lowers the EA of both forward and backward. So how do we know when we implement your method. When it states in 'practice' rather then in theory?

[shinny]

What does what coblin said have to do about the Ea of forward and backward reactions? He said side-reactions involve the consumption of the catalyst (such as say...nitrogen poisoning the V2O5 catalyst of the contact process)

EDIT: Whoops didn't see the question that all this was associated with @_@

[Collin Li]

Haha, yeah. I agree with C for that question.

[Rosie]

oh ok, this question was from one of the tsfx booklets that i got from the lectures and i guess they got the first question wrong. So everyone agrees that the first question is C.

[Rosie]

Another question:
Q. considering the reaction: Hg_(l) + 0.5I_2(g) ----> HgI_(g) 

i.) what is the Ea for the reaction 0.5Hg_(l) + 0.25I_2(g) ----> 0.5HgI_(g)?
ii.) what is the Ea value for the reaction 2HgI_(g)  ------> 2Hg_(l) + I_2(g)?

thanks

yeah the delta H for this reaction is +138kJ/mol and the activation energy for the reaction is 298 kJ/mol. There were lots of parts to this question that's why i forgot to add things in.

[Rosie]

anyone please?