Exam: http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2008chem2-w.pdf
can someone help me with the following questions:
Q7: No idea
Q8: can you explain why the concentration of hydrogen gas is higher than Iodine? and why is it that the concentration of HI increase more than all the products?
Q12: Still massively struggling with pH. Someone save me please, how can I get better at pH stuff?
Q16: I chose C but idk why its D?
Q18: Chose A, answer is D. Isnt it in galvanic cells, the negative electrode = oxidation and positive electrode = reduction?
Q19: the last three options all produce 2 moles of CO2 per mole of the fuel?
Question 7These types of questions, where they sort of combine enthalpy changes from multiple reactions, are always a bit tricky to grasp. Fortunately, we have a diagram to assist us. Looking at the diagram, going from 1Cu
2O(s) and 1/2O
2(g) to 2CuO(s) releases 312 - 170 = 142 kJ mol
-1. Now, I want to go the other direction and I also want to double the moles: 4CuO(s) --> 2Cu
2O(s) + O
2(g). So, the enthalpy change for the new reaction is
+142 x 2 = +248 kJ mol
-1. The answer is 'A'.
Question 8Well, the answer is 'B': the concentrations instantaneously increase when the volume is decreased and there is no further change in concentration as the system cannot oppose this change due to the fact that it has the same number of particles on the reactant and product sides. Your questions here are actually irrelevant to answering the question. The only reason hydrogen is more concentrated than iodine in this particular scenario is because there's more hydrogen in the mixture, and the only reason the concentration of hydrogen iodide increases more is again because it is more concentrated. We haven't been given the initial concentrations, but I can discern this from the correct graph.
Question 12Only 'II' is true, so the answer is 'A'.
'I' is incorrect because adding a base increases the pH.
'III' is incorrect because although propanoic acid is a weak acid, it will still have ionised to a very small extent.
'II' is a correct statement. Think about it: the more NaOH you add, the more you ionise the propanoic acid. Now, propanoic acid and NaOH would react in a 1:1 ratio, as propanoic acid donates one proton per molecule (i.e. is monoprotic) and sodium hydroxide only has one hydroxide ion per molecule. So, when 100 mL of 0.08 M NaOH is added to 100 mL of 0.16 M propanoic acid, the mole of propanoic acid is double so only half the amount of NaOH needed to neutralise the acid has been added. This means that only 50% of the acid has ionised at this point, so the concentration of propanoic acid = the concentration of its conjugate base.
Question 16When determining reactions using the electrochemical series, I always encourage my students to circle all the possible reactants.
So, for the electrolysis of molten NaF, and I'm assuming with inert electrodes:
- the potential oxidants are: Na
+(l), and therefore the cathode/reduction reaction will be Na
+(l) + e
- --> Na(s)
- the potential reductants are: F
-(l), and therefore the anode/oxidation reaction will be 2F
-(l) --> F
2(g) + 2e
-For the electrolysis of
aqueous NaF:
- the potential oxidants are: Na
+(aq) and H
2O(l)
- the potential reductants are: F
-(aq) and H
2O(l)
Being higher up on the series than Na
+(aq), and lower down on the series than F
-(aq), H
2O(l) is now the stronger oxidant and the stronger reductant and therefore both the cathode and anode reactions will have changed.
The answer is 'D'.
Question 18In a galvanic cell, the reductant is oxidised at the anode, and the anode is negative in a galvanic cell.
- In a cell of P and Cu, Cu is the negative electrode so it is a stronger reductant than P
- In a cell of Cu and Q, Q is the negative electrode and so it is a stronger reductant than Cu
- In a cell of Cu and R, R is the negative electrode and so it is also a stronger reductant than Cu
- In a cell of Q and R, R is the negative electrode and so it is a stronger reductant than Q.
So, the order of weakest to strongest reductant is: P, Cu, Q, R (i.e. the answer is 'D').
Question 19It's not simply a matter of looking at which equation has the greatest mole of CO
2, as the question has asked for the greatest amount of CO
2 per coulomb of electrical charge. So, the mole of CO
2 in each equation needs to be divided by the mole of electrons!
- For reaction methanol: 1/6
- For reaction ethanol: 2/12 = 1/6
- For reaction ethane: 2/14 = 1/7
- For ethane-1, 2-diol: 2/10 = 1/5
As the mole of CO
2 is only being divided by '5' for ethane-1, 2-diol, this has the greatest mole of CO
2 per coulomb of charge and therefore the answer is 'D'.