VCAA Unit 4 2008 Sample Exam

[jeremykleeman]

I just did the VCAA sample exam for the new course (http://www.vcaa.vic.edu.au/vce/studies/chemistry/2008chem2-samp.pdf)

And I found another site where there are unofficial solutions (http://www.cea.asn.au/VCAA%20Sample%20Exam%202%20solutions.pdf)

But I have a few queries as I am not sure the answers are completely correct.

MULTIPLE CHOICE

Question 2
The OCl- ion acts as a base in water according to the equation
OCl-(aq) + H2O(l) --> HOCl(aq) + OH-(aq)
When two drops of 5.0 M NaOH are added to an equilibrium mixture of OCl- in water at constant
temperature
A. the pH of the solution decreases.
B. the concentration of OH- increases.
C. the concentration of HOCl increases.
D. the ratio [OCl-]/ [HOCl] [OH-] increases.

I'm actually not sure what the right answer is here. I thought D initially but answers say B. Can someone explain?

SHORT ANSWER

Question 2
e) iii. Hydrogen can also be produced by electrolysis of a dilute solution of potassium chloride.
Give the overall equation for this reaction.

I thought it would be 2Cl-(aq) + 2H2O(l) --> Cl2(g) + H2(g) +2OH-(aq) going by the electrochemical series but answers say 2H2O(l) → O2(g) + 2H2(g)

Also check out Question 3 c) i) 1. and Question 5 b) and c) (they confused propan-1-ol with propane)
I am pretty sure answers are wrong

What does everyone think?
Thanks,
Jeremy

[Collin Li]

If you add NaOH to a solution, the OH- concentration will increase. Le Chatelier's principle applies, but note that the principle only partially opposes the change, so overall, the OH- concentration still remains higher (even if the reaction goes backward to try to get rid of it).

BTW, option D is just the reciprocal of the acidity constant. The acidity constant will not change by adding NaOH. Only temperature will change this.

[Collin Li]

I thought it would be 2Cl-(aq) + 2H2O(l) --> Cl2(g) + H2(g) +2OH-(aq) going by the electrochemical series but answers say 2H2O(l) → O2(g) + 2H2(g)

The potassium chloride is a trick. Look out for the oxidation reaction involving water, right below the chlorine equation. Both the water equations are more likely to react (water is a stronger reductant than chloride ions are). Just combine those two half-equations together to get that reaction.

[jeremykleeman]

Thanks for that

[bec]

Another quesiton about this....
in question 16, don't they have the of an exothermic reaction as positive? It says that "11.7 kJ of energy is released" (energy released = exothermic, yeah?) and then all the multiple choice alternatives are positive...

[riadnicolas]

I agree that 3ci is wrong - has anyone else found that? I got "deeper green," since an increase in volume should shift the position of the equilibrium to the side with more moles, which is the RHS.


Also, in question 16, don't they have the [tex] \Delta H[/tex[ of an exothermic reaction as positive? It says that "11.7 kJ of energy is released" (energy released = exothermic, yeah?) and then all the multiple choice alternatives are positive...

increasing the volume will decrease the concentrations of everything and i think chateliers principle applies it will never be as green as before. just a guess

[bec]

yep haha I just realised that and deleted that post literally just then, before you posted that. Thanks though!

Any ideas about my other question?

[riadnicolas]

Any ideas about my other question?

dont stress about it