Chemistry 3/4 2013 Thread

[lzxnl]

VCE chemistry skims through a lot of things, and as you're not introduced to anything deeper, you don't realize the complexity of chemistry that lurks beneath.
And well...calculus works as an explanation, so why not use it?

[Nato]

quick question:

need to find 2.33g AgNO3 in 398ml of water, in mol L^-1

thanks guys

[psyxwar]

quick question:

need to find 2.33g AgNO3 in 398ml of water, in mol L^-1

thanks guys

Work out the number of moles of AgNO3 first (n=m/M)

then n(AgNO3)/0.398L = molarity of soln

[lolipopper]

calculus and integration is a total bitch. 

[Jeggz]

calculus and integration is a total bitch. 

You do realise this is the Chem thread right?

[lzxnl]

You do realise this is the Chem thread right?

You do mean VCE Chem thread? Chemistry, in the form of thermochemistry, can use a lot of calculus. It's ugly. And scary.

[clıppy]

A rate of reaction multiple choice, hoping that someone can explain it.

Two thermochemical equations representing reactions involving the molecules P, Q, R and S are shown below
P_(g) -> Q_(g) + R_(g) ; ΔH = +56 kJ mol^-1
P_(g) + R_(g) -> + S_(g) ; ΔH = +45 kJ mol^-1

What would be the enthalpy change for the reaction described by the chemical equation Q_(g) + 2R_(g) -> S_(g)
A. -101 kJ mol^-1
B. -11 kJ mol^-1
C. +11 kJ mol^-1
D. +101 kJ mol^-1

[Edward21]

A rate of reaction multiple choice, hoping that someone can explain it.

What would be the enthalpy change for the reaction described by the chemical equation Q_(g) + 2R_(g) -> S_(g)
A. -101 kJ mol^-1
B. -11 kJ mol^-1
C. +11 kJ mol^-1
D. +101 kJ mol^-1

Ouch, I'm with you on this one! I'd like to see the explanation Quite difficult

[clıppy]

The book doesn't have much explanation for it apart from the answer being B

[lzxnl]

It's not rate of reaction; enthalpy changes don't directly tell you about reaction rates.

Reverse the first reaction to get R + Q => P , dH = -56 kJ/mol

Add this to the second equation to get P + Q + 2R => P + S, dH = -11 kJ/mol

The Ps cancel, so you have the reaction you want. B.

[psyxwar]

A rate of reaction multiple choice, hoping that someone can explain it.

What would be the enthalpy change for the reaction described by the chemical equation Q_(g) + 2R_(g) -> S_(g)
A. -101 kJ mol^-1
B. -11 kJ mol^-1
C. +11 kJ mol^-1
D. +101 kJ mol^-1

I got B.

Q+2R -> S, delta H = +45 kJmol-1
 
Let's say Q+2R is 0 kJmol-1, S is 45 kJmol-1 (so deltaH works)

P = Q+R, deltaH = +56kJmol-1

Sub in P = Q+R into Q+2R -> S
Q+R+R -> S

P and R's enthalpy combined is 0, but P is now Q+R which is +56kJmol-1. LHS enthalpy is now +56kJmol.

RHS (or S) enthalpy is +45kJmol1.

deltaH = 45-56=-11
deltaH = -11kJmol-1

[clıppy]

Ahh okay. Thanks psyxwar and nliu!

[Edward21]

Question, for an exothermic reaction, with its energy profile: I know the little rise is the energy absorbed to break the bonds in the reactants as the activation energy required, over the sort of turning point and down towards the level of the products is the delta H, heat of reaction. However what about that little bit above the level of the reactants on the right hand side, right after the turning point? What exactly is that? It can't have an impact upon ΔH, is the same amount of energy put in as the Ea lost, and then further energy is lost to make a negative ΔH??

http://studentweb.usq.edu.au/home/W0099066/images/Equilibrium/exothermic.png

[lzxnl]

Let's think of it this way. Energy has to be continuous, so for the energy to drop from that of the activated complex (the peak) to that of the reactants, it has to drop through every energy value on the way.

In the activated complex, we may very well have partial bonds forming. Let's take the example of reacting methyl bromide with hydroxide ion to form methanol. Here, the activated complex has carbon forming partial bonds to three hydrogens, a bromine and an oxygen, i.e. five partial bonds. However, as the carbon-bromine bond gradually weakens and as the carbon-oxygen bond gradually strengthens, the overall energy starts to drop. Eventually, the bromine atom effectively drops off (if the collision is successful) and the oxygen atom forms a bond with the carbon, in which the reaction is complete. This is what is happening when the energy is dropping.

[Edward21]

Let's think of it this way. Energy has to be continuous, so for the energy to drop from that of the activated complex (the peak) to that of the reactants, it has to drop through every energy value on the way.

In the activated complex, we may very well have partial bonds forming. Let's take the example of reacting methyl bromide with hydroxide ion to form methanol. Here, the activated complex has carbon forming partial bonds to three hydrogens, a bromine and an oxygen, i.e. five partial bonds. However, as the carbon-bromine bond gradually weakens and as the carbon-oxygen bond gradually strengthens, the overall energy starts to drop. Eventually, the bromine atom effectively drops off (if the collision is successful) and the oxygen atom forms a bond with the carbon, in which the reaction is complete. This is what is happening when the energy is dropping.

Wow, thanks so much. Btw is it true that we only need to know the selected (nitric, sulfuric or ammonia) industrial chemical for the SAC, and it is not an examinable by VCAA? It's not clear in the study design for 2013-2016