ALS3A - Section 3/ Energetics / Suggested solution/ 1 SKH...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Section 3/ Energetics / Suggested solution/ SKH Bishop Baker Secondary School A-Level Chemistry Section 3 Energetics 80 IB Q.6 (a) (i) The term is lattice energy. [1] The lattice energy of M + Cl - (s) is the standard enthalpy of formation of one mole of crystal lattice M + Cl - (s) from M + and Cl - in the gaseous phase . [2] [If one mole not mentioned, deduct 1 mark; if gaseous phase not mentioned, deduct 1 mark; do not penalise if M + Cl - (s) or standard are not mentioned.] (ii) M + (g) + e - + Cl(g) H at for Cl 2 (g) electron affinity of Cl M + (g) + ½ Cl 2 (g) + e - M + (g) + Cl - (g) ionization energy lattice energy of M of M + Cl - M(g) + ½ Cl 2 (g) H at of M(s) (accept any suitable diagram) M(s) + ½ Cl 2 (g) H f for M + Cl - (s) M + Cl - (s) [2] Deduct ½ mark for each mistake or omission (including states). Do not penalise if standard is omitted. (iii) (I) The basic assumptions are as follows : (1) The crystal lattice is composed of discrete spherical ions. [1] (2) Charge is distributed evenly around each ion. [1] (Point charges give one mark only.) (II) Purely ionic bonding exists in M + Cl - (s) and thus the assumption in (I) hold for M + Cl - (s). [1] The simple purely ionic model is not satisfied and thus the above assumptions do not hold for AgCl(s) because there is distortion of electron cloud on anion Cl - by cation Ag + to introduce covalent character in AgCl(s) . [1] (III) M belongs to group I. [ ½ ] M is very electropositive and so there is a large difference in electronegativity between M and Cl. [ ½ ] 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Section 3/ Energetics / Suggested solution/ [If correct group is given but wrong reason, e.g. because M forms a +1 ion give no mark. Correct group but no reason ½ mark; If mention that this is a theoretical case and therefore M cannot be placed in the Periodic Table 1 mark.] (i) H 1 = - n w c θ n x , a n w = number of moles of water; c = molar heat capacity; θ = change in temperature; n x , a = number of moles of anhydrous salt X H 1 = - (100/18)(75.6)(8.0) 8.0/160 = - 672 00 J mol - 1 = - 67.2 kJ mol - 1 [1] [if no sign no mark; if wrong sign no mark.] (ii) H 2 = + n’ w c θ n x , h where n x , h = number of moles of hydrated salt X H 2 = +(100/18)(75.6)(1.0) (13. 0/250) = +8077 J mol - 1 = + 8.077 kJ mol - 1 [1] [if no sign no mark; if wrong sign no mark.] (iii) H 1 X(s) + large quantity of water → X(aq) H 3 + yH 2 O(l) H 2 (accept any suitable X . yH 2 O(s) + large quantity of water diagram) [1 or 0] (iv) H 3 = H 1 - H 2 = (-67.200) - (+8.077) = -75.277 kJ mol -1 [1] (v) The possible assumptions are as follows : (1) The thermal capacity of the container is negligible. (2) The molar heat capacity of solution of X is similar to water.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 08/17/2011 for the course CHEM 100 taught by Professor Fleet during the Fall '08 term at Oxford University.

Page1 / 13

ALS3A - Section 3/ Energetics / Suggested solution/ 1 SKH...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online