# The 10 hardest questions we've ever asked in VCE Chemistry

There’s no doubt that getting plenty of practice with exam questions is the key to acing any subject, VCE Chemistry included. And it’s the hard questions that truly test whether you know what’s going on or not. In that respect, ConnectApp has got you covered. Having done dozens of past exam papers ourselves, we know the kinds of questions that often trip students up, and they’re the exact type of questions we’ve designed for ConnectApp. Below are some of our favourite, sadistic, student-trapping questions from across Unit 3 and 4 VCE Chemistry.

## 1. When Universal Gas Equation meets Specific Heat Capacity.

A scientist burns 0.005 L of propane at SLC to heat 250.0 grams of water, and finds that the temperature of the water increases by 0.4020°C. The chemical equation for the combustion of propane is given below:

C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g)

Using this data, the enthalpy change for the above reaction would be:

1. +2081 kJ mol-1
2. -2081 kJ mol-1
3. +3697 kJ mol-1
4. -3697 kJ mol-1

### Question analysis

We are given the volume of propane at SLC (big hint that we should use the Universal Gas Equation!), as well as the mass and temperature change of water (big hints that we should use the Specific Heat Capacity formula).

Our goal is to find the enthalpy change for the given reaction – since the units for enthalpy change are kJ mol-1, that means we need to somehow calculate an energy value (in kJ) and the moles of propane, and then divide the two. Now we were saying before that the data in the questions hints at us to use the Specific Heat Capacity formula and Universal Gas Equation – so how do they help us find energy and n(propane)? Well, we can use the mass of water and its temperature change in the SHC formula to get the energy released from burning propane, and we can use the volume of propane and the conditions at SLC to find n(propane) from the Gas Equation!

### Working out

Using the SHC formula to find the energy released from burning propane (remember to convert the energy to kJ because that’s what’s used for ΔH!):

Using the volume of propane and conditions at SLC (100 kPa, 298 K) in the Universal Gas Equation, we can find n(propane):

Now that we have both the energy released and the moles of propane, we can divide the two to find the ΔH value. Note that we should always include a + or - sign in front of ΔH to indicate whether the reaction is endothermic or exothermic, respectively. In this case, the reaction is exothermic because the temperature of the water increased when propane reacted (indicating it overall released heat to the surroundings).

## 2. Manic galvanic reaction predictions!

A galvanic cell consists of one half-cell that is composed of an inert graphite electrode in a solution containing 1.0 M Sn4+(aq) and 1.0 M Sn2+(aq).

A suitable second half-cell that could be used so that the polarity of the electrode in this second half-cell is positive is:

1. A silver electrode in a 1.0 M solution of silver nitrate.
2. A tin electrode in a 1.0 M solution of Sn2+(aq).
3. An iron electrode in a solution of 1.0 M Fe3+(aq) and 1.0 M Fe2+(aq)
4. An inert graphite electrode in a solution of 1.0 M Br-(aq)

### Question analysis

We already know for certain that one of our half-cells is the Sn4+(aq)/Sn2+(aq) half-cell. What we’re trying to figure out is what we should connect it to so that the other half-cell has a positive electrode. The word ‘positive’ gives us a huge hint about what reaction will be occurring in each beaker. In galvanic cells, the positive electrode is the cathode, where reduction takes place. Thus, the Sn4+(aq)/Sn2+(aq) half-cell must be the negative anode, where oxidation takes place. This allows us to turn to our electrochemical series in the Data Book and draw a big fat arrow going backwards (to indicate oxidation) for the Sn4+(aq)/Sn2+(aq) half-cell. Since we want the Sn4+(aq)/Sn2+(aq) half-cell to be the oxidation half-cell, we’ll need to react it will something in the upper left of the electrochemical series.

### Working out

For electrochemistry questions, the electrochemical series is your best friend – to get the most use out of it, use a pencil to make annotations on it, like circling species and drawing arrows (visualising what’s happening really helps you see what species are going to react).

Here, we can draw an arrow going backwards for Sn4+(aq)/Sn2+(aq) because we know this half-cell is the place of oxidation. For a spontaneous reaction, we need a species in the top-left to react with our Sn2+(aq). Therefore, any of the species in the large orange box will do, but of all the multiple choice options, the only half-cell that fits this criteria is the Ag+(aq)/Ag(s) half-cell.

## 3. Don’t let this fuel cell fool you.

Consider the following half-equations and their standard electrode potential:

CO2(g) + 4H+(aq) + 4e- ⇌ CH2O(g) + H2O(l), Eo = -0.43 V

CO2(g) + 6H+(aq) + 6e- ⇌ CH3OH(g) + H2O(l), Eo = -0.39 V

2CO2(g) + 12H+(aq) + 12e- ⇌ CH3CH2OH(g) + 3H2O(l), Eo = -0.33 V

O2(g) + 4H+(aq) + 4e- ⇌ 2H2O(l), Eo = +1.23 V

O2(g) + 2H+(aq) + 2e- ⇌ H2O2(aq), Eo = +0.68 V

Two of the above half-equations are used in the fuel cell shown below:

The reactants that would have to enter at positions A and B, respectively, to achieve the given voltage are:

1. CH3OH(g), O2(g)
2. O2(g), CH3OH(g)
3. CH3CH2OH(g), O2(g)
4. O2(g), CH3CH2OH(g)

### Question analysis

We’re given some half-equations with their Eo values, and told that two of them are used in the fuel cell shown. Looking at this fuel cell, the only information it shows us is the electron flow, and that the voltage is 1.62 V. How are these useful to us? Well, the electron flow will help us put the oxidant and reductant in the correct positions (either position A or B), while the voltage will help us pick our two half-equations.

### Working out

Recall that predicted cell voltage is given by the following formula:

Eo(cell) = Eo(higher half-cell) - Eo(lower half-cell)

With a bit of experimenting (taking higher Eo values and subtracting lower Eo values from them) we can figure out that the following two half-equations give us the desired Eo:

O2(g) + 4H+(aq) + 4e- ⇌ 2H2O(l), Eo = +1.23 V

CO2(g) + 6H+(aq) + 6e- ⇌ CH3OH(l) + H2O(l), Eo = -0.39 V

Eo = +1.23 - (-0.39) = 1.62 V

Now we just need to determine which reactant will go where. Since the half-equation with the oxygen gas has the higher Eo, this will be the reduction reaction. Therefore, the methanol will be oxidised. Looking at the diagram, the only clue we’re given about where to place the two reactants is the direction of the electron flow. Electrons are flowing into electrode A, so the reduction reaction must be occurring here (since reduction is gain of electrons). Thus, oxygen gas will react here. Conversely, electrons are flowing out of electrode A, so oxidation must be occurring here. Hence, methanol will react here.

## 4. Make Michael Faraday proud.

In a classroom experiment, a solution of nickel (II) hydroxide (Ni(OH)2) was electrolysed. The following results were obtained:

• Current: 12.0 A
• Time: 150 seconds
• n(gas) formed: 0.01 mol

From these results, what is the experimental value of one Faraday (F) that the student would obtain from their results?

1. 90 000 C mol-1
2. 180 000 C mol-1
3. 45 000 C mol-1
4. Cannot be obtained without the number of moles of nickel formed.

### Question analysis

We’re told that a solution of nickel (II) hydroxide is undergoing electrolysis, and we’re given the current, time, and the moles of gas formed. Based on this, we’re asked to find the experimental value of Faraday’s constant. A great way to approach these kinds of questions involving Faraday’s laws is to first write out the two equations, then circle the variable we’re trying to find (F in this case), and place ticks next to variables we know.

As you can see, for us to be able to find F, we’re going to first have to work out what Q and n(e-) are.

### Working out

Since we’re given the I and t, we can figure out Q using Q = I x t:

Finding the n(e-) is slightly more challenging: we’re given the n(gas) being formed from the electrolysis of nickel (II) hydroxide, so we can get the n(e-) the molar ratio between the e- and the gas being formed… that means we need to figure out what this gas is! Just like any other electrolysis question, to predict products formed we turn to our electrochemical series: we circle all the species present in the question, and identify the highest thing on the left (strongest oxidant) and the lowest thing on the right (strongest reductant). Therefore, Ni2+(aq) ions will be reduced to Ni(s), and OH-(aq) ions will be oxidised to O2(g) and water. So, the mystery gas must be O2!

As we can see from the half-equation, the number of moles of electrons will be four times greater than the number of moles of oxygen gas produced.

Now that we have Q and n(e-), to find the experimental value of one Faraday, we just need to rearrange Q = n(e-) x F, and sub in the values we calculated:

## 5. Equilibrium calculation requiring a sneaky assumption.

Hydrogen iodide decomposes according to the following equilibrium reaction:

2HI(g) ⇌ H2(g) + I2(g), K = 0.0333 at 1100 K

Some hydrogen iodide gas is placed in an empty 10.0 L vessel. At equilibrium, it is found that the concentration of HI is 0.55 M.

The concentration of hydrogen gas at equilibrium would be closest to:

1. 0.010 M
2. 0.10 M
3. 0.275 M
4. Cannot be determined from the information provided.

### Question analysis

We’re told that hydrogen iodide is added to an empty vessel, and that at equilibrium, its concentration is 0.55 M. We’re already given K, and we’re asked to find the concentration of H2.

We know of two methods for solving equilibrium calculations:

• Rearranging the K expression, and
• Using ICE.

At first it seems like ICE is the way to go because we don’t know [I2]. It turns out that we can’t actually use ICE because we’re not given any initial amounts/concentrations! So, we’re going to have to rearrange the K expression to get [H2] by itself… but what about [I2]?

Well, since HI has been added to an empty vessel, we know that the concentrations of both H2 and I2 are initially zero. As HI begins to decompose, the concentration of H2 and I2 will increase at the same rate. Therefore, at equilibrium, these two species will have the same concentration. This means that we can replace [I2] with [H2] to simplify our equation, and proceed to solve it.

## 6. The caveat to Le Chatelier’s Principle.

Colourless dinitrogen tetroxide and brown-coloured nitrogen dioxide exist in equilibrium according to:

N2O4(g) ⇌ 2NO2(g), K = 4.5 M at 80oC

Both reactants are placed in a reaction vessel and allowed to establish equilibrium. Then, the volume of the reaction vessel is then halved.

Select the correct statement. Compared to the initial equilibrium, at the new equilibrium:

1. The moles of NO2 will have increased, and the mixture would be darker.
2. The moles of NO2 will have decreased, and the mixture would be darker.
3. The moles of NO2 will have increased, and the mixture would be lighter.
4. The moles of NO2 will have decreased, and the mixture would be lighter.

### Question analysis

We’re told that we have an equilibrium system of N2O4 and NO2, and that its volume is halved. The question wants us to figure out how the amount of NO2 and colour of the mixture at the new equilibrium position will compare to the initial position, so this is a classic application of Le Chatelier’s Principle. Remember, according to this principle, whenever you make a change to an equilibrium system, it’ll always try to oppose the change, but it’ll never be able to fully oppose the change.

### Working out

Decreasing the volume of the vessel will increase the pressure inside the vessel (because there’s now less room for the particles). Therefore, the system will want to decrease pressure by favouring the side with fewer particles. In this case, there is one particle on the reactants side, and two particles on the products side, so the backwards reaction will be favoured. Therefore, the number of mol of NO2 will decrease.

However, this does not mean the mixture will become lighter. Since the system can’t fully oppose the volume decrease, the concentration of NO2 will still be higher than it was initially, as you can see from the diagram below. And since colour intensity depends on concentration, the mixture will be darker. This often tricks students on the exam, so make sure you remember that the system can never fully oppose the change made to it.

## 7. Reaction pathways with compounds you’ve never seen before.

Benzene is a very useful organic compound that can have various functional groups attached to it to serve different functions. For example, consider the incomplete reaction pathway below:

The structure that should go in the blank box is:

### Question analysis

We’re given an incomplete reaction pathway involving some derivatives of benzene, and we’re asked to choose the structure that should go in the blank box.

Even though we haven’t learnt about reactions involving benzene, we can still figure out what the initial structure should be by looking at what’s happening with the functional groups. What we can see is that some compound has reacted with Cr2O72- to form an aldehyde (we know this because the middle structure has a CHO group). The aldehyde has then also reacted with the Cr2O72- to form a carboxyl group (COOH). The fact that we have Cr2O72- here is a huge hint that these are oxidation reactions (remember that Cr2O72- and MnO4- are two really common oxidising agents). From what we’ve learnt about organic chemistry, we know that it is primary alkanols which undergo oxidation to form aldehydes, and then carboxylic acids. So, what we need to do is work backwards to turn the aldehyde into the alcohol again, making sure we keep the number of carbon atoms the same.

### Working out

We know that the starting compound needs to be a primary alkanol (contains an OH group), so we can straightaway eliminate options B and D (because B is a ketone, and D is an aldehyde). Now we have to pick between A and C, which differ by one carbon atom.

The tricky thing here is that the reaction pathway has been represented with skeletal formulae, which can make it difficult to ‘see’ where the carbon atoms should be. When compounds are drawn this way, carbon atoms are represented as being the ends or corners of lines, and hydrogen atoms are not always drawn (but we can determine the number of hydrogen atoms by counting how many covalent bonds carbon has already made, bearing in mind that it can make a maximum of four bonds). It can be useful to draw in the carbons and hydrogens to get a better understanding of what’s going on:

Looking at the structure of the second and third compound, we can see that there needs to be a carbon atom coming off the ring, which then has the OH group attached to it. Therefore, we can eliminate option A (since this has OH attached directly to the ring), leaving us with option C.

## 8. The ultimate NMR detective question.

An unknown compound is isolated from a reaction mixture. After analysis of its composition, it is found that the molecular formula of the compound is C3H6O2. To determine the structure of the compound, 1H-NMR spectroscopy is used, producing the following spectrum:

The IUPAC name of the molecule that produced the spectrum is most likely:

1. Propanoic acid
2. Propanal
3. Methyl ethanoate
4. Ethyl methanoate

### Question analysis and working out

We’re told that the molecular formula of an unknown compound is C3H6O2, and we’re given its 1H-NMR spectrum. Using this, we’re asked to figure out the identity of the compound.

A good place to start is to think about what compounds are possible. This will narrow down our options and also help us focus on certain data to collect. The compound we’re working with has two oxygen atoms, which is a major clue because only the carboxyl and ester functional groups have two oxygen atoms. One of the multiple choice options provided is propanal, but since this only has one oxygen atom, we can straightaway eliminate it.

Now let’s turn to the spectrum. To avoid getting lost in data, it’s good to approach the 1H-NMR spectrum systematically in the following order:

1. Number of peak clusters (tells you number of environments).
2. Relative peak areas (tells you the number of H atoms in each environment).
3. Peak splitting (tells you about the number of H atoms in adjacent environments, so helps you figure out what environments are next to each other).
4. Chemical shifts (different environments produce different chemical shifts, so these can be used to identify the environments; however, do this as a last resort because looking them up in the Data Booklet can be time-consuming).

We can see that there are three peak clusters, which means that there are three hydrogen environments in the compound. By quickly sketching the structures of propanoic acid, ethyl methanoate, and methyl ethanoate, we can see that methyl ethanoate only has two hydrogen environments, so couldn’t have produced the spectrum.

We now have to decide between propanoic acid and ethyl methanoate. If we consider the relative peak areas, we can see that the three different hydrogen environments have 1, 2, and 3 hydrogen atoms. Looking at propanoic acid and ethyl methanoate, we can see that they also have environments with 1, 2, and 3 hydrogen atoms, so this doesn’t provide any new insight. The peak splitting for propanoic acid and ethyl methanoate would also be the same because the environments are positioned very similarly: both have a CH3 next to a CH2, and other lonely hydrogen environment that won’t split at all.

Since we’ve exhausted all the other data, the only thing we can do now is look up the chemical shifts in the Data Booklet. One method is to look up the chemical shifts of the peaks and see what environments they could correspond to; the other method is to think of what unique environments propanoic acid and ethyl methanoate have and look up what chemical shifts these environments have. Since we know we have to choose between propanoic acid and ethyl methanoate, the latter method is a more efficient one. The key difference between propanoic acid and ethyl methanoate is that the former has a carboxyl group, whereas the latter has an ester, so these are the environments we should look up.

Comparing these chemical shifts to the ones in the 1H-NMR spectrum, there’s no way that the compound could be propanoic acid because there’s no peak that’s greater than 9.0 ppm. Also, our spectrum has a peak at 4.2 ppm, which corresponds to the chemical shift of the two hydrogens in COOCH2, which reinforces the fact that it is ethyl methanoate that produced the spectrum.

## 9. A common misconception about how strong/weak acids/bases react in volumetric analysis.

Consider the following pH curve for the titration of an aliquot of ethanamine, a weak base, with hydrochloric acid (HCl), a strong acid.

Suppose the same titration was repeated, except that the ethanamine solution was replaced with a sodium hydroxide (NaOH) solution that has the same concentration as that of the ethanamine solution.

Select the correct statement:

1. The same volume of HCl would be required to reach the equivalence point in both titrations, but the pH at the equivalence point for the NaOH/HCl titration would be higher than that for the CH3CH2NH2/HCl titration.
2. The same volume of HCl would be required to reach the equivalence point in both titrations, but the pH at the equivalence point for the NaOH/HCl titration would be lower than for the CH3CH2NH2/HCl titration.
3. A greater volume of HCl would be required to reach the equivalence point in NaOH/HCl titration because NaOH is a stronger base than ethanamine, and its equivalence point would occur at a higher pH.
4. A greater volume of HCl would be required to reach the equivalence point in NaOH/HCl titration because NaOH is a stronger base than ethanamine, and its equivalence point would occur at a lower pH.

### Question analysis

In this question, we’re essentially asked to compare two titrations – one involving ethanamine and the other with sodium hydroxide (NaOH). And what we’re specifically asked to compare is the volume of HCl required, and the pH at the equivalence point. Here’s where we have to be super careful: a lot of students get confused because ethanamine is a weak base while NaOH is a strong base. In volumetric analysis, this just affects the shape of the titration curve and at what pH the equivalence point occurs. On the other hand, the volume of reactants required will only depend on the concentrations of the reactants. You’ll see this more concretely in the working out below.

### Working out

Ethanamine and sodium hydroxide solutions both have the same concentration. That means the aliquots (volumes) of ethanamine and NaOH both have the same number of moles of base! To determine the amount of moles (and therefore the volume) of HCl that’s required to react with each aliquot, we need to consider the mole ratios in the chemical equations:

CH3CH2NH2(aq) + HCl(aq) → CH3CH2NH3+(aq) + Cl-(aq)

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Since both ethanamine and NaOH combine in a 1:1 ratio with HCl as seen in the reactions above, this means each titration would consume exactly the same number of moles of HCl. Therefore, the same volume of HCl would be required.

Although the volumes of HCl are the same, the pH values at the equivalence point for each titration will differ because this depends on the strength of the acid and base. Since NaOH is a strong base and HCl is a strong acid, the pH at the equivalence point will be around 7 (diagram A). On the other hand, ethanamine is a weak base, so the pH at the equivalence point is skewed towards acidity – a pH lower than 7 (diagram B). Hence, the titration with NaOH will have an equivalence point at a higher pH than that with ethanamine.

## 10. An ace question combining multiple organic chemistry concepts.

The human body regulates its blood pressure a number of ways. One of these is by the action of a peptide hormone called angiotensin II, which causes blood vessels to constrict (become narrower), thus increasing blood pressure.

Angiotensin II is created when the Angiotensin-converting enzyme (ACE) hydrolyses a bond in a polypeptide called angiotensin I.

The interaction between angiotensin I and the active site of ACE is shown below:

Which of the following statements about the interaction between angiotensin I and ACE is correct?

1. A salt bridge forms between the deprotonated carboxyl group of isoleucine and protonated amino group of ACE.
2. The R groups of phenylananine, histidine, and leucine interact with the active site of ACE via hydrophobic forces.
3. The function of ACE is to catalyse the hydrolysis of a peptide linkage in Angiotensin I, creating Angiotensin II and a tripeptide: Phe-His-Leu.
4. The ACE active site includes a Zn2+ ion, which is classified as a cofactor because it is required for ACE to be able to perform its catalytic function.

### Question analysis

We’re given a diagram of the interaction between angiotensin I and ACE, and we’re asked to pick out the correct statement about what’s occurring.

Since the statements use names of amino acids, it’s a good idea to first look up the structures in the Data Booklet and figure out what amino acids are involved. Then you can start thinking about the interactions happening. Remember that there are many types of interactions that can take place between a substrate and enzyme, but these depend on the nature of the R groups – that is, whether they’re polar/non-polar, and what functional groups/atoms they have.

Since this is the type of multiple choice question that asks us to select the ‘correct’ statement, the best way to go about it is to carefully read each statement and see if any word or concept is incorrect.

### Working out

Statement A: Here we should ask ourselves the following:

• Is there a salt bridge?
• Is it forming between the COO- of isoleucine, and the NH3+ of an amino acid in ACE?

Looking at the diagram, we can see there is a salt bridge between the NH3+ of an amino acid in ACE and an amino acid. Now we just need to check if that amino acid is isoleucine.

Looking at the Data Booklet carefully, the structure of isoleucine has the central CH bonded to CH. However, the diagram in this question has the central CH bonded to CH2. Therefore, it’s not isoleucine! In fact, the amino acid is leucine. This is a common error to make because the structures of leucine and isoleucine are so similar (in fact, they’re structural isomers of each other – hence the name isoleucine). Therefore, Statement A is incorrect.

Statement B: Here we should ask ourselves:

• Are the amino acids shown phenylananine, histidine, and leucine?
• Do they interact via hydrophobic forces?

Looking at the Data Booklet carefully, we can work out that the amino acids are indeed Phe, His, and Leu. Now the tricker part: what intermolecular forces do they have?

To answer this, we need to consider whether the R groups of these amino acids are polar or non-polar. Both Phe and Leu have only carbon and hydrogen in their R groups, which means they must be non-polar. Therefore, they will interact via hydrophobic forces.

However, histidine has nitrogen atoms in its R groups, so it is polar. This means it will interact with ACE via dipole-dipole interactions, making Statement B incorrect.

Statement C: Here we should ask ourselves:

• Is ACE hydrolysing a peptide linkage?
• Are the products Angiotensin II and a tripeptide?

To answer the first dot-point, recall that peptide linkages comprise the atoms CONH. From the diagram we can see there’s a squiggly line right in the middle of a CONH group, which means ACE is indeed hydrolysing a peptide linkage.

In regards to the second dot-point, we know that Angiotensin II will be one product (because the background information in the question tells us this), but will we get a tripeptide? If we look at the amino acids to the right of the hydrolysed peptide linkage, there are only two! Therefore, a dipeptide of His and Leu will form, making Statement C incorrect.

Statement D: Here we should ask ourselves:

• What is a cofactor?
• Is Zn2+ acting as a cofactor according to the diagram?

Recall that the definition of a cofactor is “a non-protein chemical that is required for an enzyme to function”. Therefore, to determine if this statement is correct or not, we need to determine if Zn2+ is required for ACE to function.

Looking at the diagram, we can see that the C=O group in the hydrolysed peptide linkage is interacts with Zn2+, suggesting that Zn2+ is indeed required for the process to occur. Therefore, we can safely classify Zn2+ as a cofactor.