Kinetics - Modify Te 07 Acj H1 Ans

(Chemical Kinetics and Chemical Bonding)

Suggested Answers 1 (a) (i) Comparing Expt 1 and 2: Increasing [propanone] by four times causes rate to increase by four times. Reaction is first order with respect to propanone. Comparing Expt 1 and 4: Increasing [H+] by two times causes the rate to increase by two times. Reaction is first order with respect to H+. Comparing Expt 3 and 4: Increasing [iodine] by 8 times did not cause the rate to increase. Reaction is zero order with respect to I2. Since the concentration of the intermediate is directly affected by the concentration of propanone and the concentration of intermediate is directly affected by the concentration of H+. Thus, the rate equation of the slow step which is first order with respective to the intermediate, will also be first order with H+ and propanone. This is shown in the slow step of the reaction. Thus, the mechanism is plausible. Rate = k[H+][propanone]

(iii)

Using experiment 1: 0.001 = k(0.01)(0.01) k = 10 mol-1 dm3 min-1

(i)

Energy

Ea (slow step)

Ea (step 3)

OH

Ea (step 1)

(b)

(ii)

O H3 C

C

+

H2C

C

CH3

OH H3C

+H CH

C

+

CH3

O IH 2 C

3

C

+

ΔHr CH3 HI

Reaction Coordinate

Note: Only need to indicate Ea of the slow step and the ∆Hr. I have included the other Ea so that you can see that step 2 has the highest Ea. ∆Hr = +22 kJ mol-1 (step taken: bond broken – bond formed) (ii)

H+ acts as a homogenous catalyst. Kwok YL

(Chemical Kinetics and Chemical Bonding)

Number of molecules

At temperature, T

0

ƒ ƒ ƒ (c)

(i)

ƒ ƒ

(ii)

ƒ ƒ

ƒ ƒ

H+ provides an alternative reaction pathway with a lower energy barrier. Therefore, there will be more reactant molecules which possess the minimum energy for the reaction to occur. Hence, the number of effective collision increase and hence rate of reaction increases.

The bond strength is in the order of Cl–Cl being the strongest and I–I being the weakest. This is because as we move from Cl to I, the atomic radius increase. Since formation of covalent bond is due to overlapping atomic orbital. Hence, the overlapping of atomic orbitals between two I atoms is less effective than the overlapping of atomic orbital between two Cl atoms, since the former is larger than the latter.

Melting point trend: Cl2 < Br2,< I2 ƒ

ƒ

Energy

From the Data Booklet: Br–Br, Cl–Cl, and I–I ƒ

(d) ƒ

Ea(catalysed) Ea(uncatalysed)

Are all simple discrete molecules. Melting these substances would require the breaking of van der waals’ forces of interaction between them. Since, I2 has the largest electron cloud size, it is the most polarisable and hence its induced dipole moment is the largest. Therefore, its id-id interaction is the strongest, thus it has the highest melting point. Cl2 has the smallest electron cloud size and it is thus the least polarisable and hence its induced dipole moment is the smallest. Therefore, its id-id interaction is the weakest, thus the lowest melting point.

Rate of iodination of propanone occurs as fast as the rate of chlorination of propanone. As the reaction occurs in a multiple steps reaction, the rate of the reaction is dependent on the activation energy of the slow step. This is because that the breaking of the halogen-halogen bond does not happen in the rate determination step. The activation energy for the slow step for iodination of propanone and the chlorination of propanone is the same. Thus rate of reaction remains the same despite I–I being weaker than the Cl– Cl.

Kwok YL