Help! Chemistry MCQ
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Q1) What is the correct order for the enthaply of combustion from the least exothermic to the most exothermic?
But-1-ene, trans-but-2-ene, cis-but-2-ene
Q2) Ethanal reacts with CN- from HCN in the presence of a weak base.
In a similar reaction, -CH2COCH3 ions are generated when CH3COCH3 reacts with a strong base. Which one of the following compounds is the product when ethanal reacts with -CH2COCH3?
A. CH3CH(OH)CH2COCH3 B (CH3)2C(OH)CH2CHO
C (CH3)2C(CHO)CH2OH D (CH3)2C(OH)COCH3
Q3) Which of the following processes are both the enthalpy change H and S are positive.
A C2H4(g) + H2(g) -->C2H6(g)
B H2O(s) ---> H2O(g)
C H2O2(l) -->H2O(l) + 1/2O2(g)
D NH4NO3 (s) + aq --> NH4+(aq) + NO3-(aq)
From this question onwards, A = 1,2,3 correct, B= 1,2 , C=2,3, D=1
Q4) The reaction between X and Y is given as 2X + 2Y --> 2XY. The mechanism of the reaction is as shown below:
2X <---> X2
X2 + Y --> XY + X(radical) slow
X(radical) + Y --> XY
Which of the following statement is true?
1. Doubling the concentration of X doubles the rate of reaction
2. Doubling the concentration of X and Y increases the rate of reaction by 8 times
3. The overall order of the reaction is 3.
Q5) Which of the following are species of electrophilies?
1. Br2, NO2+
2. AlCl3, HNO3
3. Na+, (CH3)C+
Q6) Manganese (IV) oxide acts as a catalyst in the decomposition of H2O2 to O2. What alteration to the original experimental conditions would increase the volume of O2 produced over time?
1. adding more H2O2
2. lowering the temperature
3. adding more H2O
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I will give comments and suggestions, rather than work out the answer for you. As an 'A' & 'O' Levels Chemistry tuition teacher now (ex-MOE teacher), I guide my students to the correct thought processes to help them work out the answers for themselves. It's more helpful this way, generally.
Q1) What is the correct order for the enthaply of combustion from the least exothermic to the most exothermic?
But-1-ene, trans-but-2-ene, cis-but-2-ene
Suggestion :
The larger the magnitude difference between the stability of reactants and products, the more exothermic (or endothermic) the reaction.
Ask yourself which is more stable, trans or cis isomer? You should be aware that healthy cis-fatty acids are altered to form toxic, cancer-causing trans-fatty acids when foods are cooked, worst of all deep fried (google Aajonus Vonderplanitz for more info on diet and health). And the position of the double bond in alkenes, relates directly to stability of the alkene : electron donating alkyl groups stabilize the double bond in alkenes. So does but-1-ene have more alkyl groups next to the C=C, or does the cis/trans isomers of but-2-ene?
Q2) Ethanal reacts with CN- from HCN in the presence of a weak base.
In a similar reaction, -CH2COCH3 ions are generated when CH3COCH3 reacts with a strong base. Which one of the following compounds is the product when ethanal reacts with -CH2COCH3?
A. CH3CH(OH)CH2COCH3 B (CH3)2C(OH)CH2CHO
C (CH3)2C(CHO)CH2OH D (CH3)2C(OH)COCH3
Suggestion :
The carbanion nucleophile generated attacks the delta-positive carbonyl carbon in an nucleophilic addition reaction. Draw the Kekule structures and mechanism to figure out which option is the answer.
Q3) Which of the following processes are both the enthalpy change H and S are positive.
A C2H4(g) + H2(g) -->C2H6(g)
B H2O(s) ---> H2O(g)
C H2O2(l) -->H2O(l) + 1/2O2(g)
D NH4NO3 (s) + aq --> NH4+(aq) + NO3-(aq)
Suggestion :
Entropy - Compare LHS vs RHS. Gases have highest degree of disorderliness, followed by aqueous, lastly solid. When both sides have the same state, compare no. of moles.
Enthalpy - For some reactions, use the Data Booklet bond enthalpies/energies data to determine endo or exo. For others, like (B), changing of state, is pretty obvious whether endo or exo. For others, like (D), you should know that by Hess Law, Solution enthaply = (endothermic) Lattice Dissociation enthalpy + (exothermic) Hydration enthalpy (which is the ion-dipole interactions between polar water molecule and the cations & anions).
[Updated 20 Sept 08 : Regarding option 4 (dissolving or solution of ammonium nitrate into water).
This is a well known endothermic reaction (ie. your hand feels cold when holding the test tube containing the reaction mixture).
What about entropy change? This is trickier than at first glance.
Entropy would both increase and decrease (there will be of course either a net increase or decrease, overall) for the following reasons :
1) Entropy would be expected to increase because solid state to aqueous state involves an obvious increase in disorderliness.
2) Entropy would be expected to decrease because of ion-dipole interactions between water molecules and cations & anions (charge density and hence ion-dipole interactions is always higher & stronger for cations than anions, do you know why?). Disorderly water molecules now arrange themselves in an orderly fashion around the ions, particularly those of high charge density (eg. Al3+, Fe3+).
3) Entropy would be expected to decrease because of a lowered temperature (due to the reaction being endothermic), and hence less kinetic energy results in less disorderliness of all species present, ions and molecules.
Overall, entropy change would be near zero (points 1 would cancel out the effects of points 2 & 3); and without further data given, impossible to state with any certainty if net entropy change is positive or negative.
For the purpose of this particular question, by elimination, because we know option B is the obvious answer, we can deduce that (assuming this question is correct, valid and accurate) net entropy change for dissolving of ammonium nitrate is negative.
Q4) The reaction between X and Y is given as 2X + 2Y --> 2XY. The mechanism of the reaction is as shown below:
2X <---> X2
X2 + Y --> XY + X(radical) slow
X(radical) + Y --> XY
Which of the following statement is true?
1. Doubling the concentration of X doubles the rate of reaction
2. Doubling the concentration of X and Y increases the rate of reaction by 8 times
3. The overall order of the reaction is 3.
Suggestion :
- The slow step is the rate determining step in which the stoichiometry of the elementary equation will reveal the order of the reaction with regards to the reactants. (note that you can't use stoichiometry for overall equation to determine order of reaction, you can only use stoichiometry on the rate determining elementary equation to determine order of reaction).
- The formation of the free radical is usually the slow step, and the subsequent reaction of the free radical is the extremely fast step, because free radicals are highly unstable and thus highly reactive.
- If the rate determining step involves intermediates (eg. X2), we have to substitute another expression for X2 into the rate equation, because we do not want intermediates as part of the rate equation. From the 1st step (the association of two X atoms to form an X2 molecule), we obtain the expression k1[X]^2 = k2[X2] ==> [X]^2 = [X2](k2)/(k1). Substituting this expression into the rate equation of Rate = k3 [X2] [Y], we obtain the final rate equation of Rate = k [X]^2 [Y]. In other words, the reaction is 2nd order in X and 1st order in Y.
Q5) Which of the following are species of electrophilies?
1. Br2, NO2+
2. AlCl3, HNO3
3. Na+, (CH3)C+
Suggestions :
Electrophiles are formal positive or partial positive charged species that invite attack by nucleophiles (species that have at least one available lone pair for donation). As for diatomic molecules like Br2, there can be instantaneous or induced dipoles that cause a shift in electron density between the atoms, resulting in a delta-positive or delta-negative, the delta-positive functioning as an electrophile to be attacked by nucleophiles such as the pi-bond of an alkene.
(Read the following 2 paragraphs very carefully, it might be a little difficult for some students to understand).
Note that electrophiles must have energetically accessible orbitals for attack by nucleophiles. For instance, NH4+ is not an electrophile because being in Period 2, nitrogen does not have empty d orbitals to expand its octet, thus it cannot be attacked by nucleophiles, even if it has a positive formal charge on the nitrogen.
That being said, there can be instances where resonance will result in energetically accessible orbitals being freed up in an electrophile to enable nucleophilic attack. For instance, draw the Kekule structure of NO2+. Initially, the main resonance contributor (ie. the main Kekule structure) is O=N+=O, two double bonded oxygens with a central positive formal charged nitrogen. Although N is in Period 2 and cannot expand its octet, when the nucleophile (eg. pi-electrons of benzene ring) attacks the N, one of the pi-bonds between N and O will shift over to become a lone pair on O, resulting in a negative formal charge on that O. The final NO2 attached to the benzene ring (for instance, if this was an electrophilic aromatic substitution reaction), has 2 formal charges which cancel out : a +ve formal charge on N (because it only has 4 valence electrons from 4 bond pairs, but it is in group V; but note that it certainly has a stable octet), and a -ve formal charge on one of the Os.
Futher comments on this Qn :
1. Br2, NO2+
2. AlCl3, HNO3
3. Na+, (CH3)C+
Br2, NO2+, (CH3)C+ are hence functional electrophiles as explained above. AlCl3 functions as an Lewis acid by accepting an electron pair from a nucleophile (since the Al does not yet have a stable octet in AlCl3, and is willing to accept an electron pair from a Cl2 molecule, forming -AlCl4 and a Cl+ electrophile). The Kekule structure of HNO3 has a negative formal charge on the O, and a positive formal charge on the N. However, HNO3 or nitric acid, functions as an acid (proton donor) or a base (proton acceptor, in the case of reacting with sulfuric acid in the nitrating mixture), an also as a strong oxidizing agent (eg. with Bart Simpson adding ethanol to nitric acid), but not so much as either a nucleophile or an electrophile (note that it has both a positive and a negative formal charge side by side on the N and O, which repels incoming electrophiles and nucleophiles respectively. Moreover, if HNO3 were to behave as an electrophile, and a nucleophile successfully attacks the positive formal charged N, to avoid violating the octet rule, one of the pi-bonds with oxygen will have to become a lone pair on the oxygen, resulting in a species with too many negative formal charges to be stable; bearing in mind that being a strong acid, the acidic proton would also dissociate resulting in yet another negative formal charge for yet another O in the species, which is too unstable).
As for Na+, it has neither the inclination nor tendency to accept electron pairs - Na desperately wants to be oxidized to Na+ (enegretically stable octet), which will resist efforts to be reduced back to Na.
A carbocation has 3 bond pairs and 0 lone pairs, it certainly functions well as an electrophile. And the 3 methyl groups are electron donating by induction, stabilizing somewhat the carbocation allowing it to exist long enough as an electrophile species.
Q6) Manganese (IV) oxide acts as a catalyst in the decomposition of H2O2 to O2. What alteration to the original experimental conditions would increase the volume of O2 produced over time?
1. adding more H2O2
2. lowering the temperature
3. adding more H2O
Suggestion :
Adding more H2O2 reactant would obviously result in more O2 product over (a long period of) time.
The decomposition of hydrogen peroxide is exothermic (determine this using bond enthalpies in Data Booklet), so heat is regarded as a product; hence lowering the temperature (ie. reducing the availability of a product) will result in position of equilibirum to be shifted to the RHS, product favoured reaction, so more O2 will be produced at equilibrium, assuming this is an equilibrium reaction.
But is the decomposition of hydrogen peroxide an equilibrium reaction? One of the products, O2, is gaseous, which means that it will leave the reaction mixture, and hence pull over the position of equilibrium completely to the RHS. Hence decomposition of H2O2 is certainly *not* an equilibrium reaction. Hence, the decomposition of hydrogen peroxide would be complete (after a long period of time) anyway, and hence lowering temperature does not make a difference (over a long period of time).
Adding more water does two things - increases molarity of product water, and dilutes molarity of reactant hydrogen peroxide. In regard to kinetics, you'd expect lowerig molarity of reactant would result in a slower rate of O2 produced. And again, assuming it is an equilibirum reaction, you'd expect increasing molarity of RHS product would shift position of equilibriuim to the LHS, resulting in less O2 produced. But because the gaseous product O2 pulls the position of equilibrium completely to the RHS, hence over a long period of time, adding water doesn't affect the final volume of O2 produced (apart from having an additionally small percentage of oxygen gas dissolving into a greater quantity of water).
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Thanks for the kind advice UltimaOnline ;)
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Originally posted by Uncertain:
Thanks for the kind advice UltimaOnline ;)
Most welcome, Uncertain!
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Note : I edited my comments to the last question.
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Note (20 Sept 08)
I've edited and updated the post containing my solutions/suggestions above, as follows :
Q3) Which of the following processes are both the enthalpy change H and S are positive.
A C2H4(g) + H2(g) -->C2H6(g)
B H2O(s) ---> H2O(g)
C H2O2(l) -->H2O(l) + 1/2O2(g)
D NH4NO3 (s) + aq --> NH4+(aq) + NO3-(aq)
Suggestion :
Entropy - Compare LHS vs RHS. Gases have highest degree of disorderliness, followed by aqueous, lastly solid. When both sides have the same state, compare no. of moles.
Enthalpy - For some reactions, use the Data Booklet bond enthalpies/energies data to determine endo or exo. For others, like (B), changing of state, is pretty obvious whether endo or exo. For others, like (D), you should know that by Hess Law, Solution enthaply = (endothermic) Lattice Dissociation enthalpy + (exothermic) Hydration enthalpy (which is the ion-dipole interactions between polar water molecule and the cations & anions).
[Updated 20 Sept 08 : Regarding option 4 (dissolving or solution of ammonium nitrate into water).
This is a well known endothermic reaction (ie. your hand feels cold when holding the test tube containing the reaction mixture).
What about entropy change? This is trickier than at first glance.
Entropy would both increase and decrease (there will be of course either a net increase or decrease, overall) for the following reasons :
1) Entropy would be expected to increase because solid state to aqueous state involves an obvious increase in disorderliness.
2) Entropy would be expected to decrease because of ion-dipole interactions between water molecules and cations & anions (charge density and hence ion-dipole interactions is always higher & stronger for cations than anions, do you know why?). Disorderly water molecules now arrange themselves in an orderly fashion around the ions, particularly those of high charge density (eg. Al3+, Fe3+).
3) Entropy would be expected to decrease because of a lowered temperature (due to the reaction being endothermic), and hence less kinetic energy results in less disorderliness of all species present, ions and molecules.
Overall, entropy change would be near zero (points 1 would cancel out the effects of points 2 & 3); and without further data given, impossible to state with any certainty if net entropy change is positive or negative.
For the purpose of this particular question, by elimination, because we know option B is the obvious answer, we can deduce that (assuming this question is correct, valid and accurate) net entropy change for dissolving of ammonium nitrate is negative.