Unit 4 Questions MEGATHREAD!

[HarveyD]

i) Water is acting as the oxidant and the reducant
use the equations/formulae to find the volume of oxygen gas produced at the anode (oxidation)
i.e. use faraday's rule then pv = nRT

ii) Increasing the concentration to produce chlorine gas

Just my guess, may be wrong

[Mr. Study]

Hello!

Just a few questions.

I have attached a picture of the question.

1. The answers say its B, Why is it B? :S

2. Questions 10 and 11 refer to the following information.
The equilibrium between NO2 and N2O4 can be described by the equation: 2NO2(g) -> N2O4(g) Kc = 0.010 M-1 ; ΔH = -58 kJ mol-1

It can also be described by a second equation 1/2N2O4(g) -> NO2(g)

How would I find the value of Kc for the second equation?
(I know it has something to do with the reciprocal of the first equation).

3. A calorimeter is used to study the reaction between 100 cm3 of 0.5M HCl(aq) and 100 cm3 of 0.5M NaOH(aq). The calibration factor of the calorimeter and its contents is 140 J oC-1. The initial temperature of the calorimeter and its contents is 25.0 oC . The temperature rise during the reaction is 20.0 oC.

How would I find DELTA H? (I have no idea how to make the triangle )

Thank you very much!

[omgwtf123]

since the thermochemical equation has been reversed and halved, we apply two formulas : Kc (new)=1/Kc when the equation is reversed
                                                                                                                                              Kc(new)=(Kc)^1/2 when the equation is halved
                                                                                                                                        Kc(new)=1/(Kc)^1/2 when both are applied
                                                                                                                                        Kc(new)= 1/(0.010)^1/2
                                                                                                                                        Kc(new)= 10M^1/2
                                                                                                                                         

[omgwtf123]

3) 1cm^3= 1mL xD
find the number of mol of both substances and determine which one is excess and limiting (we use the limiting reagent, but for this case, the balanced equation for the reaction is HCl(aq)+NaOH(aq)--->H2O(l)+NaCl(s), there is no excess or limiting reagent, hence we can use both.
n(HCl)=cv
          =0.1 x0.5
          =0.05 mol
n(NaOh)=n(HCl)
              =0.05mol
Energy released= CF x delta T of combustion
                           = 20.0 x140
                           =2800J
                            =2.8KJ
delta H= (Energy released/n( HCl or NaOH)) x the coefficient of NaOH or HCL
           = (2.8/0.05) x 1
           = -56KJ/mol       
Hope my logic is correct, if not, someone please correct me, i am happy to learn

[omgwtf123]

1) Its B, because Y is the activation energy for the forward reaction, Y is the activation evergy for the backwards reaction and X would be the energy released into the environment becuase the energy profile shows an exothermic reaction. In exothermic reactions E(released)> E(absorbed).

[omgwtf123]

in my second sentence i mean X and not Y , sorry

[omgwtf123]

omg sorry, i mean Z, i think i am losing it >.<

[Mao]

edit mate, there's an edit button. =]

[Mr. Study]

Ah! Thank you very much for the answers omgwtf123.

EDIT: Damn, I hate the 1cm cube = 1ml!

[generalkorn12]

For Fuel Cells, how would you be able to tell where is the Anode/Cathode if there's no charges? Would you have to base it off of what reactants are going into the cell, then checking the electrochemical series?

[b^3]

When discharging, fuel cells are similar to galvanic cells (as in chemcial to electrical energy). If the direction of electron flow is given, you can work out that the electrons are leaving the anode and going to the cathode. If they direction of electron flow is not given, then yes you can use the electrochemical series to predict which on would be the cathode and which one would be the anode. I hope that helps, but there may be an easier way if anyone wants to add anything.

[Vincezor]

I have a question that I 'sorta' know how to answer, but am unable to express it properly (For 3 marks anyway)

Explain why the change in pH is less when 10mL of 0.10M propanoic acid is diluted to 100mL with water, than when 10mL of 0.10M hydrochloric acid is diluted to 100mL with water.

[b^3]

Ok someone will need to double check this one but here goes nothing.
HCl is a strong acid and so basical everything (close to it) ionises. i.e. we can use c1v1=c2v2
c2=c1v1/v2
=0.1*0.01/0.1=0.01M
Now Propanoic acid is a weak acid, i.e. it ionises according to an equilibrium equation.

So as we add more water, then the position of the equilibrium shifts and to compensate, more H₃O⁺ (i.e. H⁺) is produced. As more H⁺ is produced to compensate, the change in concentration is less, so the change in pH is less.
I hope that helps Vincezor, I think this is similar to a question off a VCAA exam from a couple of years ago that a lot of people got wrong.

[Christiano]

I have a question and I can't seem to get the calculation right:

Ka value of ethanoic acid is 1.75 x10^-5. A brand of vinegar had a pH value of 3.65. Calculate concentration of ethanoic acid in the vinegar.

I've gotten to this stage: 1.75 x10^5 = [H30+]^2 / [ethanoic acid]. Not sure what to do from there, I keep getting the answer wrong.

[HarveyD]

find the [H3O] using the pH given
then sub in and solve?
did you try that?