MJC H1 Chem Prelim Paper 2 (Answers)

Answers for MJC 2008 H1 Chem Prelim Paper 2 1(a) The graph below represents the first i onization energies from beryllium to strontium. Explain the general trend presented by the graph. [2] Energy Element Be Mg Ca Sr Down the group, from Be to Mg, Nuclear charge increases as proton number increases. Valence electrons are added to a next outer shell ⇒ valence electrons are furt her away from the nucleus. Significant increase in shielding effect outweighs the increase in nuclear charge. Effective nuclear charge decreases Weaker electrostatic force of attraction between nucleus and valence electrons ⇒ Ionization energy decreases (b) Write the electronic configurations of chromium ion, Cr 3+. [1] 1s 2 2s22p6 3s23p63d3 1

(c) Between Cr3+ and F -, state and explain which ion would be deflected the most when they are passed through an electric field? [2] Extent of deflection ∝ q m q m ratio for Cr3+ = 3 52 = 0.0577 q m ratio for F- = 1 19 = 0.0526 Therefore Cr 3+ will deflect the most (d) The first ionization energies of ber yllium, boron, nitrogen and oxygen are presented in the graph below. Energy Element Be B N O Explain the decrease in ionization energy from: (i) Be to B Be: 1s2 2s2 B: 1s22s22p1 ƒ The 2p electron in B is at a higher energy level than the 2s electron in Be ƒ Hence, less energy is required to re move the 2p electron from B compared to Be. 2

(ii) N to O [3] N: 1s2 2s2 2p3 O: 1s 22s22p4 ƒ There is inter-electron repulsion between the paired el ectrons in the 2p orbital of O. ƒ Hence, less energy is required to re move the 2p electron from O compared to N [Total: 8] 2 Certain enthalpy changes such as ent halpy change of formation cannot be found by direct experiment. Instead it can be found by applying Hess’ Law to simple energy cycles. (a) 500 cm 3 of water was heated in a calorimeter by burning a 1.30 g of butane, C4H10. The temperature rise of the water recorded was 30 °C. Calculate the enthalpy change of combustion of butane. [2] Heat absorbed, Q = mc∆T = 500 x 4.18 x 30 J = 62700 J No of moles of butane burnt = 1.30/58 = 2.24 x 10 -2 Heat absorbed by water, Q = Heat released by combustion of butane, Q’ Q’ = ΔHc (C4H10) x n ΔHc (C4H10) = – -2 62700 2.24 x 10 =–2799 kJ mol-1 3

(b) The energy cycle for the complete combustion of butane is shown below. C4H10 (g) + 13/2 O2(g) 4CO2(g) + 5H2O (l) ΔHc (C4H10) 4C(s) + 5H 2(g) Using your answer in part (a) and the following enthalpy changes, calculate the enthalpy change of formation of butane. Δ Hc (C) = - 393.5