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