H2 Chem MCQ Questions

gohby SG

Hi,

I have some H2 Chem MCQ questions that to discuss/solicit for opinions. Any help would be much appreciated, cheers!

Q1: Modified CJC Prelims 12:

The Thermit Reaction involves mixing iron(III) oxide with aluminium powder in a crucible, with a suitable fuse to start the reaction. The reaction is as follows:

Fe2O3(s) + 2Al(s) → Al2O3(s) + 2Fe(l)

The fuse is first ignited, where it will burn in oxygen, forming the oxide with a large release of heat required for the Thermit reaction to take place. The commonly used material for the fuse is a clean magnesium strip. Which of the following helps to explain why a strip of magnesium is suitable to be used as a fuse?

1: The numerical value of the enthalpy change of formation of magnesium oxide is very large.

2: The strip increases the surface area for magnesium to react with the oxygen at a faster rate.

Ans: 1 and 2 are correct

Remarks: I think that 1 and 2 are not precise enough to serve as explanations.

For 1, the numerical value of the enthalpy change of formation of magnesium oxide being very large does not explain its suitability, since it does not tell us anything about the sign of the enthalpy change. The enthalpy change of formation should have been large AND negative to generate enough heat to kick start the Thermit Reaction.

For 2, where is the basis of comparison to say that the strip increases the surface area for magnesium? I could argue that (compared to magnesium power), the strip is not as suitable to be used as a fuse because the surface area is lower.

Q2: AJC Prelim 2012

The Gibbs free energy change of a system determines whether a reaction is spontaneous, while the equilibrium constant indicates the extent of reaction. What does the following pair of values for a reaction system indicate?

Values:

ΔG –50.8

Kc 5.80x10^8

A: No reaction

B: Position of equilibrium lies to the left

C: Some extent of reaction

D: Reaction goes to completion

Ans: D

Remarks: I note that the reaction is spontaneous and the Kc is a very large value, but shouldn’t the answer be C, given that the reaction is still after all, a reversible reaction (and therefore it does not go to completion)?

Q3: How do I know that the following equation: H2O (l) → H+ (aq) + OH- (aq) has a positive enthalpy change of reaction?

UltimaOnline SG

Hi Gohby, it's been a couple of years since you posted discussions on H2 Chem qns.

For Q1, you're right of course. At the same time, it's not feasible to have detailed, mini-essays as options for an MCQ. Hence, your explicit points are to be taken as implied by the summarized options. The student is to select the best possible options in any case.

Edited to add : while you're right to say the option 1 could be improved to explicitly specify the formation enthalpy of MgO needs to be exothermic, nonetheless (or it could also be the question's intention to test that) the candidate should be able to draw the inference from MgO's extremely strong ionic bonds (due to high cationic and anionic charge densities) that the formation enthalpy would naturally be strongly exothermic and thus thermodynamically favourable. While Singapore JC Prelim paper questions are notoriously ambiguous and debatable, but Cambridge isn't always much better (every year there's always at least a couple of debatable questions in which JC teachers disagree on what's the best answer, and different TYS authors provide different answers), so advise your (more able) students to always psychologically profile the Cambridge question setter's intention, and choose the best available options.

PS. Check out this YouTube video of a Chemistry teacher demonstrating the Thermite reaction : https://www.youtube.com/watch?v=qqvQwfH_wGQ

For Q2, notice that the Kc value is so large, that the reaction is mathematically 99.999% complete, which when rounded off to 3 sig fig, is hence to be regarded as 100% complete. As always, choose the best available option(s).

For Q3, the forward reaction involves only covalent bond breaking, and is thus endothermic. Of course, ion-dipole bonds are formed exothermically, but are of a significantly lesser magnitude compared to covalent bonds. Even if the question were to involve the hydroxonium ion (instead of simply H+), the final answer in this context would still be the same.

gohby SG

Thank you very much, UltimaOnline. :)

I have some questions on Ideal Gases:

Q1: What is the pressure (in Pa) of a sample of hydrogen gas that has density of 8gm-3 at 300 degrees celsius?

Answer: (573x8x11.2x101)/273

Remarks: Pressure = (DensityxRxT)/Mr = (8x8.31x573)/4?? Why doesn't this work?

Q2: Which changes in conditions would not increase the voulme of a fixed mass of gas?

Pressure/Kpa Temperature/K

1. Halved Halved

2 Doubled Halved

3 Halved Doubled

Answer: 1 & 2

Remarks: According to the equation pV=nRT wouldn’t the answer be 1 only?

Q3: 1dm³ of gas X weighs 1g and 1dm³ of gas Y weighs 5g under the same conditions of temperature and pressure. Which of the following statements are correct?

The ratio of the Mr of X to Y is 1:5

The average velocity of the molecules in gas x and gas Y are the same at the same temperature

The number of molecules in Y in 1dm³ is 5 times the number of molecules of X in 1dm³

Answer: 1 only

Remarks: Isn’t the answer 1 and 2? Shouldn't the average velocity of gaseous molecules be the same at the same temperature? 3 is wrong because it should have been 1.5 times instead of 5 times.

UltimaOnline SG

Thank you very much, UltimaOnline. :)

I have some questions on Ideal Gases:

Q1: What is the pressure (in Pa) of a sample of hydrogen gas that has density of 8gm-3 at 300 degrees celsius?

Answer: (573x8x11.2x101)/273

Remarks: Pressure = (DensityxRxT)/Mr = (8x8.31x573)/4?? Why doesn't this work?

Q2: Which changes in conditions would not increase the voulme of a fixed mass of gas?

Pressure/Kpa Temperature/K

1. Halved Halved

2 Doubled Halved

3 Halved Doubled

Answer: 1 & 2

Remarks: According to the equation pV=nRT wouldn’t the answer be 1 only?

Q3: 1dm³ of gas X weighs 1g and 1dm³ of gas Y weighs 5g under the same conditions of temperature and pressure. Which of the following statements are correct?

The ratio of the Mr of X to Y is 1:5

The average velocity of the molecules in gas x and gas Y are the same at the same temperature

The number of molecules in Y in 1dm³ is 5 times the number of molecules of X in 1dm³

Answer: 1 only

Remarks: Isn’t the answer 1 and 2? Shouldn't the average velocity of gaseous molecules be the same at the same temperature? 3 is wrong because it should have been 1.5 times instead of 5 times.

Totally welcome, Ghoby :)

Q1: What is the pressure (in Pa) of a sample of hydrogen gas that has density of 8gm-3 at 300 degrees celsius?

I would advise students to work on density separately from the ideal gas formula.

Density = Mass / Volume, which implies Mass = 8 x 1 = 8g in 1 m3, which implies 4 mol of H2 gas in 1 m3.

PV=nRT, which implies P=(nRT)/V, which implies P = ( 4 x 8.314 x 573 ) / 1 = 1.9056 x 10^4 Pa.

Of course, if it's an MCQ qn, and the 4 options only show working instead of the final answer (doncha just hate these type of silly MCQs), then you've no choice but to do ur best to backward engineer and make an educated guess on (what's on) the mind of the question setter, how *he* does *his* own working to arrive at the answer.

Looking at the given answer you posted, he's using molar volume of ideal gas at stp = 22.4 dm3.

Q2: Which changes in conditions would not increase the voulme of a fixed mass of gas?

Pressure/Kpa Temperature/K

1. Halved Halved

2 Doubled Halved

3 Halved Doubled

It's a trick qn. Option 1 has no change in the volume. Option 2 decreases the volume. Either way, the volume doesn't *increase*.

Q3: 1dm³ of gas X weighs 1g and 1dm³ of gas Y weighs 5g under the same conditions of temperature and pressure. Which of the following statements are correct?

The ratio of the Mr of X to Y is 1:5

The average velocity of the molecules in gas x and gas Y are the same at the same temperature

The number of molecules in Y in 1dm³ is 5 times the number of molecules of X in 1dm³

This MCQ tests your fundamentals of Physics. Kinetic energy = 0.5 x mass x velocity^2

At the same temperature, all ideal gases have the same average kinetic energy. But because different gases have different masses, hence the average velocities of different gases are different, even as the average kinetic energy of different gases are the same, at the same temperature.

Statement 3 is incorrect, because assuming ideal gas behavior, the same volume of both gases (at the same temperature and pressure) would contain the same no. of molecules (whether mono-atomic or poly-atomic, is irrelevant in this context).

gohby SG

Thank you very much UltimaOnline.

I have further questions on Chem Bonding/Atomic Structure.

Q1: Beryllium difluoride reacts readily with trimethylamine, (CH3)3 N to form a stable addition product. Nitrogen trifluoride has no reaction with trimethylamine.

Which of the following statements are true?

The molar ratio for the reaction between beryllium difluoride and (CH3)3 N is 1:1.

Remarks: The answer shows that 1 is false, but why? Isn’t the addition product formed with a dative bond between N and Be, thereby giving N a stable octet.

Q2: What type(s) of bonding occur in NaOCl?

Van der Waals forces

Remarks: The answer shows 1 is wrong, but isn’t there Van der Waals forces in OCl-?

Q3: Which of the following statements is wrong?

A: The melting points of the Group I hydroxides increase with increasing relative molecular mass.

I know A is wrong because the melting point decreases, but I am pondering why that's the case. Is it because the ionic bond between Gp I cations and the hydroxide ions become weaker as they go down the group, given that the effective nuclear charge of the valence electrons of the bigger Gp I metals like Cs is lower, thereby increasing the covalency of the ionic bond and thereby weakening the strength of the ionic bond?

Q4: At one stage in the radioactive decay of the osope 235 92 U, the isotope 211 82 Pb is present. How many alpha particules and beta particules will be emitted in the decay? An alpha particle is 4 He²+ and a beta particle is 0 e-

2 -1

A: 2 alpha particles and 6 beta particles

B: 6 alpha particles and 2 beta particles

C: 6 alpha particles and 6 beta particles

D: 12 alpha particles and 12 beta particles

Remarks: Why is there -1 in the configuation of the electron 0 e-, as if an electron contains one missing proton? -1

I would have thought that there would be 5 alpha particles being emitted since the difference in protons between U and Pb is 10.

UltimaOnline SG

Thank you very much UltimaOnline.

I have further questions on Chem Bonding/Atomic Structure.

Q1: Beryllium difluoride reacts readily with trimethylamine, (CH3)3 N to form a stable addition product. Nitrogen trifluoride has no reaction with trimethylamine.

Which of the following statements are true?

The molar ratio for the reaction between beryllium difluoride and (CH3)3 N is 1:1.

Remarks: The answer shows that 1 is false, but why? Isn’t the addition product formed with a dative bond between N and Be, thereby giving N a stable octet.

Q2: What type(s) of bonding occur in NaOCl?

Van der Waals forces

Remarks: The answer shows 1 is wrong, but isn’t there Van der Waals forces in OCl-?

Q3: Which of the following statements is wrong?

A: The melting points of the Group I hydroxides increase with increasing relative molecular mass.

2 -1

A: 2 alpha particles and 6 beta particles

B: 6 alpha particles and 2 beta particles

C: 6 alpha particles and 6 beta particles

D: 12 alpha particles and 12 beta particles

Remarks: Why is there -1 in the configuation of the electron 0 e-, as if an electron contains one missing proton? -1

I would have thought that there would be 5 alpha particles being emitted since the difference in protons between U and Pb is 10.

No problem :)

Q4. From 235,92,U to 211,82,Pb we note that 10 protons and 14 neutrons are emitted, which has a total mass of 24, and an overall charge of 10+. The only option which fits this, is option B. About your remark : because for this question, the question specifies that beta particles are electrons (instead of positrons), and as such an electron beta particle is a particle that contains zero protons and negligible mass.

Q3. Ionic bonds weaken because cationic charge densities decrease down the Group. Degree of covalency further decreases, not increases (since the magnitude of electronegativity difference between the metal and oxygen increases). Finally, effective nuclear charge does *not* decrease down the Group, as is often wrongly taught by Singapore JC teachers. The real reason for decreasing electronegativity and decreasing endothermic ionization enthalpy down the group is due to Coulomb's law (ie. increasing atomic radius).

Q2. Van der Waals forces exist between molecules. Because sodium hypochlorite (Latin name) or sodium chlorate(I) (Stock name) is an ionic compound, the electrostatic attractions between the ions are named ionic bonds, rather than van der Waals forces. To address your question directly : even if it were a neutral molecule rather than an ion, van der Waals forces exist intermolecularly, ie. between molecules, not intramolecularly, ie. within molecules.

Q1. Because trimethylamine is a monodendate Lewis base (ie. only 1 dative bond can be donated per N atom, and only 1 N atom is present per trimethylamine molecule ; note it is incorrect to say "because the N atom has only 1 lone pair" as often wrongly taught by Singapore JC teachers, because even if 2 lone pairs were present, eg. H2O ligand, each O atom can only donate 1 dative bond because to donate both lone pairs as dative bonds would incur an unacceptably destabilizing dipositive formal charge), and because the electrophilic Be atom of the beryllium difluoride Lewis acid has only 2 existing bond pairs, therefore stoichiometrically, 2 moles of trimethylamine Lewis bases are required to nucleophilically add (ie. donate dative bonds) unto each mole of beryllium difluoride Lewis acid molecule, in order for Be to have 4 bond pairs and hence a stable octet noble gas configuration.

gohby SG

Regarding Q4, wouldn't the emission of 4 alpha particles result to the emission of 12 protons and 12 neutrons, how does that gel with the answer of 10 protons and 10 neutrons?

Regarding Q1: Thanks, I overlooked the fulfillment of the octet structure for beryllium. If I may be inquisitive about it - how do I know that there wouldn't be any steric hindrance that would prohibit the formation of the addition product, given that there are 3 methyl groups in trimethylamine?

ACJC Prelim 12:

The empirical formula of a fluorocarbon is CF2. At the same temperature and pressure, 1dm³ of the fluorocarbon weighs 8.93g while 1dm³ of fluorine gas weighs 1.80g.

What is the molecular formula of the fluorocarbon?

A. CF2,

B. C2F4

C. C4F8

D. C5H10

Remarks: I worked out the Mr of the fluorocarbon to be ~188 --> [(1.8/38) moles of fluorocarbon weighing 8.93g]. Whilst C is definitely the best answer, I have some disquiet about the choice as 188 vs 200 is a bit of a stretch.

I have found a website dated July 2000 of a similar question (http://www2.hmc.edu/www_common/chemistry/TUTORIALS/reviewSolution07Frame.html), except that the 1dm³ of fluorine gas weighs 1.70g. This will result to the Mr of the fluorocarbon being 200. Hence I am wondering if there is something wrong with this question.

UltimaOnline SG

Regarding Q4, wouldn't the emission of 4 alpha particles result to the emission of 12 protons and 12 electrons, how does that gel with the answer of 10 protons and 10 electrons?

ACJC Prelim 12:

The empirical formula of a fluorocarbon is CF2. At the same temperature and pressure, 1dm³ of the fluorocarbon weighs 8.93g while 1dm³ of fluorine gas weighs 1.80g.

What is the molecular formula of the fluorocarbon?

A. CF2,

B. C2F4

C. C4F8

D. C5H10

Regarding ur radioactive decay query, if it were a straightforward case, then yes your answer would be right (and if it were a free response question, the student should certainly write your answer). However, radioactive decay is a complex matter, and is really beyond the scope of the H2 Chemistry syllabus (thus this isn't a really fair question). The existence of positrons (not covered in the H2 Chemistry syllabus) makes it possible for protons and neutrons to interconvert during radioactive decay. As always, choose the best answer (in this context, one that fits with conservation of mass and charge).

Regarding ur BeF2 qn, methyl groups are small and pose little steric hindrance. Cambridge will set questions that are sufficiently clear cut (ie. as long as the question successfully tests students on concepts within the syllabus, the variables within the question won't be overly sadistic). If Cambridge wanted the able student to say, "I know, it's coz of steric hindrance!", then Cambridge would use obviously bulky groups such as phenyl groups, versus small groups such as methyl groups. They won't use something whose size is directly in-between, that even Cambridge examiners themselves wouldn't know for certain (if the steric hindrance present is sufficient to alter the outcome) without actually carrying out the experiment, let alone Singapore JC teachers, let alone Singapore (or International for that matter) A level Chemistry students. Also, if it were a free response question, then advise your students to cover both possibilities : eg. during electrophilic aromatic substitution, "if the major product turned out to be the ortho product, it's due to there being 2 ortho positions versus 1 para position; if the major product turned out to be the para product, it's due to steric hindrance and van der Waals repulsion between the existing bulky substituent and the bulky incoming electrophile".

Regarding ur ACJC qn, you're right. Either the ACJC teacher made a typo, or he/she thinks it's a good chance to test the student on whether they know to use the principle "always choose the *closest* option available, regardless of discrepancies".

gohby SG

Hi UltimaOnline, thank you for your response. :) I have 2 further questions on Chem Bonding.

Q1: Which of the following is most likely to be true for sodium ethanoate?

A: It is more soluble in water than sodium chloride

B: A solution of sodium ethanoate has a higher electrical conductivity than a solution of sodium chloride.

Remarks: Both choices are wrong. For A, I reckon the ion-dipole interaction between Cl- and water is more exothermic than that of CH3CO2- and water because Cl- has a higher charge density, so the ion-dipole complex is more stable. Is this correct? As for B, how do I evaluate the relative electrical conductivity of 2 ionic compounds (esp in this case where both ions in both compounds are singly charged?)

Q2: This question is regarding hydrogen bonding in ice/water.

(i) Why is the bond angle about oxygen in ice is 109.5?

Is it not possible to form the lattice arrangement if the bond angle is 105?

Wikipedia states: “This tetrahedral bonding angle of the water molecule essentially accounts for the unusually low density of the crystal lattice – it is beneficial for the lattice to be arranged with tetrahedral angles even though there is an energy penalty in the increased volume of the crystal lattice.” My question: Okay, but why is it beneficial for the lattice to be arranged with tetrahedral angles?

(ii) In my notes, it states: “Hydrogen bonding is maximised (in ice) because the water molecules are arranged in an ordered way to form a regular lattice”. My question: what is hindering hydrogen bonding in water from being arranged in an ordered way to form a regular lattice?

UltimaOnline SG

Hi UltimaOnline, thank you for your response. :) I have 2 further questions on Chem Bonding.

Q1: Which of the following is most likely to be true for sodium ethanoate?

A: It is more soluble in water than sodium chloride

B: A solution of sodium ethanoate has a higher electrical conductivity than a solution of sodium chloride.

Q2: This question is regarding hydrogen bonding in ice/water.

(i) Why is the bond angle about oxygen in ice is 109.5?

Is it not possible to form the lattice arrangement if the bond angle is 105?

No problem, keep ur qns coming :)

Q1. Yes you're right about the relative strengths of the ion-dipole interactions. Any difference in entropy effect can be taken to be negligible, unless otherwise specified by the question for A level purposes.

For A level purposes, the electrolytic strength and electrical conductivity of an ionic compound is directly proportional to the *number* of aqueous ions (regardless of magnitude of charges, eg. Na+ vs Mg2+, or charge densities, eg. F- vs Cl-) present per unit volume of solution, ie. the total *molarity* of aqueous ions. Since both sodium chloride and sodium enthanoate are completely soluble, and both dissolve to generate the same *number* of ions (per mole of ionic compound), hence electrolytic strength and electrical conductivity of both sodium ethanoate and sodium chloride are regarded as the same.

Beyond A levels, the electrical conductivity of an ionic species is also further affected by : activity coefficients, ionic atmosphere, asymmetry/relaxation effect, electrophoretic effect, electrostriction, etc.

Q2. You seem to be confusing the 105 deg bond angle *within* a water molecule, and the 109.5 deg hydrogen bond angle *between* water molecules in solid ice. Because of the molecular geometry of water (the 2 lone pairs on the O atom, and the 2 partial +ve H atoms, are tetrahedral relative to each other), and therefore ice *naturally* (hence it's "beneficial", ie. without incurring significant angle deviation or strain) takes on a tetrahedral lattice structure (we say ice is a "solid hydrogen bonded lattice structure"). The so-called slight "energy penalty" (in the Wikipedia article) refers to the work done against the surroundings when water increases its volume during the freezing process.

At temperatures and pressures of liquid water, the H2O molecules have sufficient kinetic energy to overcome the rigid hydrogen bonds that hold the H2O molecules in fixed positions in the solid ice state, and thus the H2O molecules will gladly move about as much as the temperature and pressure allows them, ie. a (thermodynamically favorable) positive entropy change (relative to the solid ice state). In the liquid state (unlike the solid and gaseous states), the hydrogen bonds between H2O molecules are continuously broken and formed, allowing the H2O molecules to move about.

So to address your question directly : it's (positive) entropy that prevents water at liquid state temperatures and pressures from remaining as a solid. Depending on the different temperatures and pressures that permits and favors water to be in the different states, it's advantageous entropy-wise (hydrogen bond breaking) for water to transit from solid to liquid to gaseous states, and it's advantageous enthalpy-wise (hydrogen bond forming) for water to transit from gaseous to liquid to solid states. Assuming constant pressure, the main determining factor (for whether water choose to be solid, liquid or gas) is hence temperature (as seen in Gibbs free energy formula).

gohby SG

Hi UltimaOnline, I have some questions on Reation Kinetics :)

Q1: http://img.photobucket.com/albums/v700/gohby/Chemistry/Q34%20ACJC_zps63m4r5pe.jpg

Response: The order of reaction wrt X is 1 and wrt Y is 1. But how to proceed from there?

Q2: 2SO2(g) + O2(g) ⇔ 2SO3 (g) ▲H = -98kJ/mol

Which of the following conditions increase both the rate of reaction and the yield of products?

Add a suitable catalyst and add a suitable solvent to dissolve SO3

Decrease the volume of the vessel

Increase the temperature and decrease pressure

Remarks: The answer is 1 and 2. I don’t understand why the answer is 1 because if we use a suitable solvent to dissolve SO3, wouldn’t the yield of sulphur trioxide drop instantly?

Q3: If the rate equation (and its concomitant rate order(s)) is experimentally derived - how come we are able to deduce it from mechanisms - isn’t that being theoretically derived then?

Q4: 2O3(g) → 3O2(g)

This reaction is thought to occur via a two–step mechanism:

Step 1 O3(g) O2(g) + O(g) fast, reversible

Step 2 O3(g) + O(g) → 2O2(g) slow

The rate law that is consistent with this mechanism is k[O3]²/[O2].

Remarks: I know how to get the answer, but I am wondering why the rate equation contains the product as well (shouldn’t it only be expressed in terms of the reactants)? How can the concentration of the product affect the rate of reaction when it is an not a reversible reaction?

Thank you very much, UltimaOnline :)

UltimaOnline SG

Hi Gohby,

For Q1, if you interpret the question to imply molecule Z to be the only product (ie. no other products formed on RHS in the overall equation), then options 2 and 3 are both possible. Option 1 is ruled out because the energy profile diagram indicates an intermediate.

For Q2, it is implied that you can subsequently retrieve your SO3 from your solvent.

For Q3, you have to experimentally derive the orders of reaction first, then use them to deduce the mechanism. Orders cannot be deduced by inspection of the overall equation, but can be deduced by inspection of the rate determining elementary step of the mechanism.

For Q4, the rate equation must be in terms of only reactants and products, but not intermediates. The product O2 can affect the rate of reaction because in the reversible elementary step, the molarity of O2 affects the molarity of O, the intermediate which is involved as a reactant in the rate determining elementary step.

You're welcome, Gohby :)

gohby SG

Hi UltimaOnline,

Thank you for your response. :)

For Q1, why does the presence of an intermediate rule out option 1 being the possible overall equation? Am I right to say any option that follows: a X + b Y → Z, where a and b are not 1 would be a plausible option?

For Q3: Am I right to say that mechanisms are merely guesses at how a reaction takes place step by step at a molecular level? Which is why when even we derive the orders from the slow step in a mechanism we are still deriving it experimentally (rather than it being theoretical since we already know the mechanism already)? At the H3 level (or even at times at H2 Chem!) one would have to deduce a mechanism from a given reaction, and deducing mechanisms requires certain theoretical knowledge (e.g. start from an electron rich region, hydride shifts etc.), so isn’t mechanisms deduced theoretically then (and thus rate equations could be said to be deduced theoretically)?

I have two other further questions:

Q5: The diagrams P, Q, R and S show how a change in conditions affects the Maxwell-‐ Boltzmann distribution of molecular energies for gas G. In each case, the original distribution is shown by a solid line and the distribution after a change has been made is shown by a dashed line. Which one of the following statements is correct?

http://img.photobucket.com/albums/v700/gohby/Chemistry/Q26%20TPJC_zpsa9xv8jje.jpg

Answer: The change shown in diagram S occurs when the pressure of G is decreased at constant temperature

Remarks: How will a decrease of pressure lead to a decrease in the total number of molecules?

Q6: Substances R, S and T react according to the following equation:

R(aq) + 2S(aq) + T(aq) → 2U(aq) + V(aq)

To find the rate equation for the above reaction, two sets of separate experiments were performed and the results are shown below.

http://img.photobucket.com/albums/v700/gohby/Chemistry/Q31%20RI_zpsurt489vr.jpg

What can be deduced from the information shown?

3 The overall order of the reaction is two.

Remarks: How do I tell whether 3 is correct?

UltimaOnline SG

Yo Gohby,

Q1. If there wasn't an intermediate, then option 1 would be true, ie. the overall reaction is simply the single (and thus rate determining) elementary step. But since there's an intermediate, option 1 could be the rate determining step, but can't be the overall equation.

Q3. Yes, you're right. For complex reactions, even after carrying out experiments to determine the order of reaction for the reactants, there may be 2 or more plausible mechanisms that fit the experimentally determined orders of reaction. For such reactions, kinetics experiments alone cannot narrow down or confirm exactly which mechanism is the correct mechanism, and further experiments (beyond the scope of A levels) must be carried out.

Q5. You're right, a decrease of pressure does *NOT* lead to a decrease in the total number of molecules (unless this was an equilibrium question with different no. of moles of gas on the LHS vs RHS). Even Cambridge A level papers have errors (not just typo, sometimes conceptual errors), let alone Singapore JC Prelim papers, which you certainly cannot trust.

Kinetic energy is directly proportional to temperature, and since temperature remains unchanged, hence average kinetic energy of the gas particles remain unchanged, and hence there would be no change to the shape of the Maxwell–Boltzmann distribution curve, and certainly no change to the total number of molecules (ie. area under the curve).

Put in another way, PV=nRT, since pressure is decreased and temperature is constant, and n is also constant (since you can't change the number of molecules unless you're talking about equilibrium reactions which is not what this question is about), it means volume must increase, which doesn't affect the Maxwell–Boltzmann distribution curve (only temperature can affect the Maxwell–Boltzmann distribution curve) for a single gas species (with a fixed mass).

Q6. Option 3 is wrong because the overall order of the reaction is 3. From inspection of the 1st graph, we deduce the reaction is 0th order w.r.t. R *and* 2nd order w.r.t. S (can you figure out why?). From inspection of the 2nd graph, we deduce the reaction is 1st order w.r.t. T (can you figure out why?).

gohby SG

Q1: Can I say that any option for a X + b Y → Z, where a and b are not 1, can be accepted (since we don’t know how does the mechanism works?)

Q5: I redacted the options but in the interest proving your conjecture here’s the question in full:

http://img.photobucket.com/albums/v700/gohby/Chemistry/Q26%20TPJC_zpszfdhvewd.jpg

Yes I was very perplexed by the choices; A-C are evidently wrong. I thought that D was an option only when I inject a sample of gas into the original sample at the same temperature (or an equilibrium system as you’ve suggested). If this has been transcribed accurately from the actual Prelim Paper, I would be worried as it would be a difficult task to read the minds of the question setters. I am surprised that even in A Levels there are examples of conceptual errors, are you able to point out a couple?

Response to Q6:

The order of reaction wrt R is 0 because with reference to the any of the graphs in the first diagram, [R] decreases linearly, i.e. a decrease in [R] does not lead to a decrease in the rate of reaction → Order wrt R = 0.

The order of reaction wrt T is 1 because the with reference to the curve in the second diagram, the half life is constant.

I wasn’t too sure about the order of reaction wrt S though, I know it can’t be 0 because from the graphs it affects the rate of depletion of R and T when [S] is increased, but here’s my take (don’t know if it’s the most efficient way of thinking): the rate of depletion of [R] is increased by a factor of 4 (this is determined by comparing the gradients of the two graphs in diagram 1) when [S] is doubled. Given that the concentrations of other reactants remain unchanged, the order of reaction wrt S has to be 2. (2²=4).

Q7: To identify an oxide of nitrogen, 0.10 mol of the oxide was mixed with 10 dm³ of hydrogen gas and passed over a heated catalyst. At the end of the reaction, 0.4 dm³ of hydrogen gas remained. The ammonia produced required 125 cm3 of 1.6 mol/dm³ HCl for neutralisation.

All gaseous volumes were measured at room temperature and pressure. What is the formula of the oxide of nitrogen?

A:NO B: N2O C: NO2 D:N2O4

Remarks: Essentially from the calculations, the ratio of nitrogen oxide: H : NH3 = 1:4:2. Hence, B is the answer because it is the only choice which fulfills this ratio (N2O + 4H2 → 2NH3 + 2 H2O). However, my qn is - how do I know that a nitrogen oxide, when mixed with a hydrogen gas will produce water (in addition to ammonia as stated in the question)? If I am unable to ascertain the other product in addition to ammonia, how do I then write an equation for all the choices to obtain the answer?

Thank you UltimaOnline :)

UltimaOnline SG

Q1. Yes, you can say that.

Q5. Yup, all 4 options are wrong.

As for Cambridge A level papers, at the very least, every year there are at least a couple of A level qns that are *ambiguous* (and *sometimes* arguably conceptually erroneous if you've higher level understanding of chemistry, but it's usually not at a level that will affect students, so not really an issue), in which Singapore JC teachers disagree amongst themselves on what the best answer should be. For examples, just buy the 3 or 4 TYS publications available on the market, and note the ambiguous questions for which different authors (many of them ex-MOE teachers themselves) provide different answers for.

Of course, Singapore JC Prelim papers are far worse in this regard. It is common... nay, invariable, that every year, a teacher from a JC will tell their students to "ignore this other JC's terrible mark scheme for this Prelim paper qn, and use our own JC's notes instead". In other words, most Singapore JC teachers won't be able to get an A grade, if they sat for another JC's prelim papers and had their papers marked by the rival JC's teachers.

Heck, it's common even *within* a JC, for teacher A to tell his class to "don't use teacher B's horrible lectures/teachings/notes/mark schemes". Hence, don't trust Prelim papers and their mark schemes.

Rest assured, that Cambridge is far more *reasonable* in this regard (compared to Singapore JC teachers and their Prelim papers and mark schemes), and as long as your students write reasonable answers in a reasonable way, they'll get reasonable marks from Cambridge.

Q6. Yes you're correct, well done. Hence it's overall 3rd order.

Q7. You're right in your calculations and deduction. Since this is an MCQ qn, students should use all available data (including options) to arrive at the best answer, as you've done. You could say this qn would be somewhat unfair to the student, if asked as a non-MCQ qn. As for your query "how do I know that a nitrogen oxide, when mixed with a hydrogen gas will produce water", this is a redox reaction (which more capable students should be able to recognize based on the description given), and by default (unless otherwise specified by the question), water is often a reactant or product of redox reactions (recall the method for balancing redox reactions).

No problem, Gohby! :)

UltimaOnline SG

From the number of views H2 Chemistry threads on this forum has garnered, it's pretty clear that hundreds of lurkers (presumably and hopefully the majority of whom are actual JC students) are interested in H2 Chemistry posts.

Well, to all you lurkers out there (especially if you're a JC student), here's a suggestion : Why don't *you* go ahead and post your own H2 Chemistry queries here.

Just register an account with SgForums, and feel free (it's literally free... of course, asking for free online help here... or for that matter anywhere... isn't as effective as actually joining my tuition, but it's still better than having no one reliable to ask for help... or if you disagree with your own private tutor or school teacher and wish to have a 2nd opinion that's reliable and trustworthy) to ask anything relevant to H2 Chemistry here.

For instance, if you wish to discuss a TYS qn or a JC Prelim qn (eg. if you don't understand or disagree with the answer given by the TYS author, or the prelim paper's mark scheme, or your school teacher's explanation. etc), you can just specify the question (eg. "2014 RJC P3 Q2" or "TYS 2014 P3 Q2" etc) and ask your question. If it's a non-TYS qn or non-Prelim qn that involves a diagram, you'll have to use your handphone to take a photo, and upload the image to your Facebook, Twitter, blog, etc, and link to the image in your post.

Terms & Conditions : Only genuine and reasonable questions from genuine and reasonable students asked in a genuinely and reasonably respectful manner will be replied to.

gohby SG

Hi UltimaOnline, I have some further questions (on ionic equilibria this time):

Re Q7 @ 21 Jul 15: this is a redox reaction (which more capable students should be able to recognize based on the description given)

Remarks: How can I recognise that it’s a redox reaction? Just because there’s reduction when a nitrogen oxide is reduced to ammonia? But how does that ensure oxidation will occur?

Q1:

http://img.photobucket.com/albums/v700/gohby/Chemistry/CJC2012_zpsnohp6cyi.jpg

Answer: 1 and 2 are correct.

Remarks:

For 2, what is meant by the point when the slope of the curve is at maximum at its centre?? Is it the point as indicated by the arrow (which I have inserted) in the diagram? If so then 2 is correct because the first neutralisation reaction would have been complete and the salt produced has no net charge.

For 3, is it wrong because at pH 9.7, there are equal concentrations of H3N+CH(CH3)COO- and H2NCH(CH3)COO- (because pH= pKa → max buffer capacity → where the concentrations of the weak acid and its conjugate base is the same)

Q2: Does this statement “H+ (aq) + OH¯ (aq) ions react exothermically” give a correct explanation for “The ionic product of water, Kw increases with increasing temperature”?

Remarks: I am of the opinion that the first statement gives a correct explanation for the second statement. Given the water dissociation equation: H2O ↔ H+ + OH¯, if H+ (aq) + OH¯ (aq) ions react exothermically, the forward reaction must be endothermic. Hence, when the temperature increases, by LCP, the equilibrium will be shifted to the right to minimise the disturbance, which would increase the concentration of H+ + OH¯ at equilibrium, which will increase the ionic product of water. However, according to the answer, the statement statement “H+ (aq) + OH¯ (aq) ions react exothermically” does not give a correct explanation for “The ionic product of water, Kw increases with increasing temperature” and I don’t understand why.

Q3:

http://img.photobucket.com/albums/v700/gohby/Chemistry/CJC20122_zps4ftdajiy.jpg

Remarks: Why is B wrong? I thought the degree of dissociation, which is related to the Ka, will only be affected by changes in temperature? How is D the correct answer? (Kb of ammonia is larger than water → it is a stronger base than water → what’s the relation of the strength of HA in liquid ammonia?

Thank you for your help, much appreciated. :)

UltimaOnline SG

Q7 - Oxides of N have OS of N in the positive OS (since O is more electronegative than N). NH3 has N in the negative OS (since N is more electronegative than H). Hence N is reduced, and when somebody's reduced, somebody's gotta be oxidized. Here, obviously H is being oxidized (since H is a reactant with OS of 0) to H2O.

Q1. Yes, you're correct. The point referred to is the 1st equivalence point.

Q2. An ambiguous qn. Don't trust Singapore JC Prelim papers. You can argue the relevancy of 1st statement vis-à-vis the 2nd statement either way. Your thought process is fine, just that the 1st statement rather *indirectly* relates to the 2nd statement. The given answer's option (which you didn't reveal) probably relates more *directly* to the 2nd statement.

Q3. The Ka of an acid depends on both temperature and solvent. By default, the solvent is always water, unless otherwise specified. Which is why, based on the solvent as water, temperature is thus the only (other) factor that affects the Ka value. Since NH3 is a stronger base than H2O, NH3 is able to deprotonate HA to a greater degree, and hence the Ka value of any acid HA, would be larger in NH3 than in H2O.

You're welcome to keep your questions coming, Gohby! :Þ

UltimaOnline SG

Btw Gohby, this thread is getting too long, please start a new thread each time you ask new questions, with the title reflecting the topics involved. This is for efficiency of viewing by members and visitors. Thanks.