Exam QNS +ANS Flashcards
RXN KINETICS
With reference to the graph above, what is the order of rxn w.r.t. Br-?
- Graph of conc. against time is a downward sloping straight line.
- Since gradient is constant, rate is independent of [Br-] Hence, order of rxn w.r.t. Br- is zero.
RXN KINETICS
Which is the rate determining step?
rate = k [NO]^2[H2]
1) 2NO -> N2O2
2) N2O2 + H2 -> N2O + H2O
3) N2O + H2 -> N2 + H2O
Step 2 is the r.d.s. 1 molecule of N2O2 intermediate and 1 molecule of H2 react in step 2 and the N2O2 intermediate is formed from 2 molecules of NO in step 1. This is consistent with the rate equation
Explain why the covalent bond formed between C1 and C2 in compound P is stronger than that formed between C3 and C4 in compound K.
The sp2 hybridised orbital of C2 atom has higher s-character, hence less diffused. The head-on overlap ( bond) between sp2 hybridised orbital of C2 and sp3 hybridised orbital of C1 becomes more effective. Hence, the covalent bond formed between C1 and C2 is stronger.
Suggest an explanation for the difference in the rate of formation of compound L.
The tertiary radical formed from compound L is more stable than primary and secondary radical due to the presence of more electron donating groups bonded to the electron deficient carbon, hence L undergoes free radical substitution at a faster rate.
By considering the entropy and enthalpy changes, explain why in practice, reaction 2 occurs more spontaneously than reaction 1 to obtain iron.
∆Gʅ = ∆Hʅ − T∆Sʅ
For reaction 1, Hʅ is positive, Sʅ is positive hence −T∆Sʅ is negative. Reaction is feasible when ∆Gʅ will be negative and this is only possible at high temperature where |H| < |T∆S|
For reaction 2, since Sʅ is approximately zero, −T∆Sʅ is approximately zero. As Hʅ is negative, ∆Gʅ will be negative (equals approximately ‒24.8 kJ mol‒1) at all temperature.
Hence, it’s more likely for ∆Gʅ for reaction 2 to be more negative than ∆Gʅ for reaction 2 hence reaction 2 is likely to occur more spontaneously.
mp bp between ionic compounds
MgCl2, AlF3 and MgCO3 have a giant ionic structure held together by strong ionic bonding between the cations and anions. L.E. q qr r MgCl2 has a cationic charge of 2+ and anionic charge of 1‒ and AlF3 has a cationic charge of 3+ and anionic charge of 1‒. The Al3+ ionic radius is smaller than Mg2+ and F‒ has a smaller radius than Cl‒.
As charge is a more important factor, the magnitude of lattice energy of MgCl2 is lower, the ionic bond in MgCl2 is weaker than AlF3 and require less energy to break. Thus, the melting point of MgCl2 is lower than AlF3.
Predict the relative proportions of J and K that are likely to be produced from 2-methylpropane. Explain your answer.
J : K = 1 : 9 There are 1 and 9 available H atoms in two different chemical environments that can be substituted to form J and K respectively. [1] Accept if structure drawn showing Ha and Hb
When the bromination is carried out and the products analysed, it is found that the mole ratio of J : K formed is about 9 : 1. Suggest two reasons for the difference between this ratio and the one you gave in (a)(iii).
- the formation of J involves a tertiary radical which contains 3 electron donating alkyl groups, whereas the formation of K involves a primary radical which contains only 1 electron donating alkyl group.
- The presence of more alkyl groups will stabilise the electron deficient tertiary radical more than the primary radical. Hence the formation of J is more favoured. [1]
- The formation of J is more exothermic (less endothermic), hence more energetically feasible/ favoured to be formed. [1]
Justify your predictions in (a)(i), by explaining why the first ionisation energy of each element you have plotted is higher/lower/similar compared to the element to its left on the graph.
As has higher first ionisation energy than Ge because it has more protons and hence a higher nuclear charge than Ge. The shielding effect experience by the outermost electron in Ge and As is approximately the same as the outermost electron is added to the same outermost shell. The effective nuclear charge is stronger in As hence more energy is needed to remove the outermost electron. [1]
Se has a smaller first ionisation energy than As.
As: [Ar]3d104s24px14py14pz1
Se: [Ar]3d104s24px24py14pz1
In Se, the outermost electron to be removed is from the paired 4px orbital. Hence it experiences interelectron repulsion due to 2 electrons occupying the 4px orbital. [1]
Rb has a much smaller first ionisation energy than Kr.
Rb has more protons and hence a larger nuclear charge than Kr. But it also has one more electron shell than Kr and hence experiences higher shielding effect. The outermost electron in Rb is less attracted to the nucleus. Hence, less energy is needed to remove the outermost electron in Rb. [1]
Explain what is meant by a weak monoprotic acid with reference to CyH.
A weak monoprotic acid is an acid that dissociates partially to donate only one proton per acid molecule. [1]
Explain, with the aid of an appropriate equation, why the pH at end-point B is greater than 7.
Cy +H2O⇌CyH+OH [1]
Cy hydrolyses in water to produce excess OH ions. [1]
Explain, with the help of an equation, how an aqueous mixture of CyH and Cy ̄ can control pH when relatively small amounts of acid is added to the solution.
When a small amount of acid is added, it is removed by a large reservoir of Cy- ions from the salt in the buffer and the pH of the solution remains almost constant. [1]
Cy-(aq) + H+(aq) CyH(aq) [1]
Explain why the reaction only starts when it is exposed to bright light.
Light provides the energy required for the homolytic fission of the Br-Br bond to generate the bromine radicals to start the reaction. [1]
Suggest why the reaction continues with increasing rate after the brief exposure to the bright light has stopped
The brief exposure to bright light produces bromine radicals that initiate chain reactions of the propagation step to produce more radicals [1] for reaction to proceed. The exothermic reaction results in an increase in temperature [1], hence rate of reaction increases.
Explain why the reaction in (b) to produce the organic halide is not preferred.
Multi-substituted products / isomeric products will be formed, resulting in low yield of the desired organic halide.
