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? Cis
isomers have bulky side chains which suffer from steric or van der
Waals repulsion. 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+). For the NH4+ ion, each H atom is capable
of forming extra strong (since the positive formal charge on N is
strongly electron withdrawing by induction) hydrogen bonds with the
solvent water molecules, decreasing entropy.
3) Entropy would also be expected to decrease slightly
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 entropy change would have to take into
consideration both the simultaneous increase and decrease in
entropy (as discussed above).
But because we know all nitrate(V) compounds and all ammonium
compounds are soluble (which means of course, ammonium nitrate is
certainly soluble!), meaning that the solution (dissolving) is a
feasible & spontaneous process, so by Gibbs Free Energy formula
(delta G = delta H - T delta S), we surmise that the entropy change
must be positive (ie. favourable) and large enough to overcome the
unfavourable (endothermic) enthalpy change.
Hence, both options B and D are correct.
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.
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).
