MJC H2 Chem Prelim Paper 3 (Answers)

Answers for MJC 2008 H2 Chem Prelim Paper 3 1(a) Hydrazine, N 2H4, has an ammonia-like odor, and is derived from the same industrial chemistry processes that ma nufacture ammonia. In some cases, hydrazine behaves like ammonia in chemic al reactions. For example, hydrazine is able to form a product when it reacts with boron trifloride, BF 3, in a molar ratio of 1:1. (i) Draw a ‘dot-and-cross diagram’ to s how the electronic structure of a molecule of N2H4. Use the electron repulsion t heory to pedict the shape of this molecule and state its bond angle. [2] N H xx xxxHN H xx H • Bond angle is compressed to 107° and N2H4 is trigonal pyramidal about central N atom. (ii) Explain why h ydrazine form a product with BF3 when they react in a molar ratio of 1:1. [1] In BF3, B is electron deficient with only 6 electrons around it. Hence, B can accept the 1 lone pair of electrons from N in N 2H4 via dative bonding to attain a stable octet configuration. (iii) Draw a diagram to show the likely shape of the product formed. [1] N N H H H H B F F F 1

(b) Hydrazine is used as rocket fuel and to prepare gas precursors used in air bags. Approximately 260 thousand tonnes of hydrazine are manufactured annually. Liquid hydrazine undergoes combustion according to the following equation: N 2H4(l) + O2(g) → N2(g) + 2H2O(l) A chemist conducted an experiment to det ermine the standard enthalpy change of combustion of hydrazine. In the ex periment, 0.210g of hydrazine was burnt as fuel to heat up a beaker containing 200 cm 3 of water. The te mperature of water rose by 4 oC. You may assume the process has 80 % efficiency. (i) Explain what is meant by standard enthalpy change of combustion of hydrazine. [1] Standard enthalpy change of combustion ( ΔHc θ) of hydrazine is the energy released when one mole of t he hydrazine is completely burnt in oxygen at 298K and 1 atm. (ii) Calculate the standard enthalpy change of combustion of hydrazine. [2] Amount of heat absorbed by water, Q = 200 x 4 x 4.18 = 3344 J Amount of heat released by reaction, Q’ = Q / 0.8 = 4180 J No of moles of N2H4 = 6.56 x 10-3 Standard enthalpy change of combustion of hydrzaine = - 6.37 x 102 kJ mol-1 2

(iii) Given the following data: enthalpy change of formation of steam = - 242 kJ mol-1 enthalpy change of vapourisation of water = + 44 kJ mol-1 and using the value y ou have calculated in b(ii), draw an appropriate energy cycle to determine the standard enthalpy of formation of hydrazine. [3] N 2 (g) + 2H2 (g) N2H4 (l) ∆Hf(N2H4) ∆Hc(N2H4) 2 x ∆Hf(H2O(g)) N 2(g) + 2H2O (g) N2(g) + 2H2O (l) 2 x ∆Hv(H2O) By Hess’ law, ∆H f (N2H4) = 2(-242) – (-637) – 2(+44) = + 65 kJmol-1 3

The standard enthalpy change of formation of hydrazine gas is +235 kJ mol-1. (i) U sing appropriate data from the Data Booklet , draw an energy level diagram to calculate the average bond energy of N-H bond in hydrazine. [3] 2N (g) + 4H (g) N 2H4 (g) N2 (g) + 2H2 (g) ∆Hf (N2H4) = + 235 kJmol-1 B.E(N-N) + 4 x B.E(N-H) 1 B.E(N ≡N) + 2x B.E(H-H) 1 By Hess Law, 160 + 4 x B.E(N-H) + 235 = 994 + 2(436) B.E (N-H) = 368 kJ mol -1 (ii) Suggest a reason for the difference in the N-H bond energy value obtained from (c)(i) with the value given in the Data Booklet. [1] The bond energy values obtain ed from the Data Booklet are average values and would differ from the experimental values. 4

(d)(i) When sodium thiosulphate is react ed separately with bromine and iodine, different products are formed. Using t he following data, and data from the Data Booklet, describe and explain the difference. S 4O6 2- + 2e ƒ 2S2O3 2- E θ = +0.09V 4SO2 + 4H+ + 6e ƒ S4O6 2- + 2H2O E θ = +0.51V SO4 2- + 4H+ + 2e ƒ SO2 + 2H2O E θ = +0.17V [3] Oxidising power of Br 2 is stronger than I2. E cell = + 0.98V > 0 ⇒ Reaction is feasible. Hence, bromine oxidizes S 2O3 2- to S4O6 2-. Ecell = + 0.56V > 0 ⇒ Reaction is feasible. Hence, bromine oxidizes S 4O6 2- to SO2. Ecell = + 0.90V > 0 ⇒ Reaction is feasible. Hence, bromine oxidizes SO 2 to SO4 2- . E cell = + 0.45V > 0 ⇒ Reaction is feasible. Hence, iodine oxidizes S 2O3 2- to S4O6 2-. Bromine can oxidize S 2O3 2- to SO4 2- while iodine can only oxidize S2O3 2- to S4O6 2-. (ii) Describe and explain why the trend of boiling points of hydrogen halides differs from the trend of thermal stability of hydrogen halides down Group VII. [3] [Total : 20] Thermal stability decreases from HCl to HI while the boiling point increases from HCl to HI Down the group, Bond length of H-X increases Bond strength of H-X decreases Hence, bond dissociation energy of HX decrease Down the group, The molecular size of hydrogen halide molecules increases. The greater the extent of distortion of the electron cloud, resulting in stronger van der Waal’s forces of attraction between the molecules. Hence, more energy is required to ov ercome these stronger van der Waals’ forces. 5

2 An example of a gas-phase reaction is the decomposition of nitryl chloride NO2Cl. At 700K, NO2Cl decomposes according to the following equation. 2 NO2Cl (g) → 2 NO2 (g) + Cl2 (g) The initial pressure of NO 2Cl is 0.0524 atm. The rate of the reaction is followed by measuring the partial pressure of Cl 2 as it changes with time, t. In such as an experiment, the partial pressure of Cl2 increased as follows. Time, t /s 0 300 900 1500 2000 3000 Partial pressure of Cl2 x 10-2 /atm 0 0.48 1.18 1.68 1.94 2.30 (a) (i) When the reaction is complete, the partial pressure of Cl 2 is 0.0262 atm. Plot a graph to show how the par tial pressure of Cl2 changes during the first 3000s of the reaction and determine the order of reaction with respect to NO 2Cl. Hence, write a rate equation for the decomposition of NO2Cl. [4] PCl2 / 10-2 atm 0 0.5 1 1.5 2 2.5 0 500 1000 1500 2000 2500 3000 3500 1.31 t1/2 1.97 t1/2 Time / s Since half-life at 1050s and is a cons tant , the order of reaction with respect to NOCl 2 =1 (ii) Calculate the rate constant for the above reaction, stating its units. [1] k = 1/2 ln 2 t = ln 2 1050 k = 6.60 x 10-4 s−1 = 6

(iii) At a certain temperatur e, the decomposition of NO2Cl follows a two-step mechanism as shown: Step 1 : NO 2Cl → NO2 + Cl Step 2 : NO 2Cl + Cl → NO2 + Cl2 The enthalpy change for the over all process is -15 kJ mol -1. For the first step, the activation energy for the forward reaction is 35 kJ mol -1 and that of a reverse reaction is 25 kJ mol-1. The activation energy for the reverse reaction of the second step is 35 kJ mol -1. Draw a labelled energy profile diagram based on the above given data. [2] Energy Ea (2nd step) = 10 kJ mol-1 Ea’ (revserse) (1st step) = 25 kJ mol-1 Ea (1st step) = 35 kJ mol-1 NO2Cl NO2 + Cl2 Ea (reverse) ’ (2nd step) = 35 kJ mol-1 ∆Hrxn = -15 kJmol-1 Reaction pathway 7

(iv) The rate of decomposition of NO 2Cl is found to increase when temperature is raised. With the aid of a sketch of th e Maxwell-Boltzmann distribution curve, explain how a small increase in temperature can lead to a large increase in the rate of decomposition. [3]