17.5 exam questions

1 a ) Use the data in the table to calculate the enthalpy of lattice dissociation of potassium oxide, K2O.

(2 × 90) + 248 + (2 × 418) - 142 + 844 = - 362 + Lattice enthalpy of dissociation Enthalpy of lattice dissociation = (+) 2328 (kJmol-1)

7 a ) Write an equation for the process that has an enthalpy change equal to the electron affinity of chlorine.

Cl(g) + e- → Cl-(g)

e) A theoretical value for enthalpy of lattice dissociation can be calculated using a perfect ionic model. The theoretical enthalpy of lattice dissociation for silver fluoride is +870 kJ mol-1. (i) Explain why the theoretical enthalpy of lattice dissociation for silver fluoride is different from the experimental value that can be calculated using a Born-Haber cycle.

Experimental lattice enthalpy includes covalent interaction but the Theoretical lattice enthalpy value assumes only ionic interaction

5 (b) The temperature of the water decreased to 14.6 °C. Calculate a value, in kJ mol−1, for the enthalpy of solution of potassium chloride. You should assume that only the 50.0 g of water changes in temperature and that the specific heat capacity of water is 4.18 J K−1 g−1. Give your answer to the appropriate number of significant figures.

Heat taken in = m × c × ΔT = 50 × 4.18 × 5.4 = 1128.6 J Moles of KCl = 5.00 / 74.6 = 0.0670 Enthalpy change per mole = +1128.6 / 0.0670 = 16 839 J mol-1 = +16.8 (kJ mol-1)

c) Gaseous methanol from this reaction is liquefied by cooling before storage. Draw a diagram showing the interaction between two molecules of methanol. Explain why methanol is easy to liquefy.

Hydrogen bonding is a strong enough force to hold methanol molecules together in a liquid

1 b) Explain why the enthalpy of lattice dissociation of potassium oxide is less endothermic than that of sodium oxide.

K+ is bigger than Na+ ion so the Electrostatic attraction between oppositely charged ions is weaker

6 c) Explain why there is a difference between the hydration enthalpies of the magnesium and sodium ions.

Magnesium ion has a higher charge density compared to sodium ions and so it attracts the oxygen in water more strongly.

5 (d) Explain why your answer to part (c) is different from the lattice enthalpy of dissociation for magnesium chloride

Magnesium ion is smaller than the calcium ion Therefore, it attracts the chloride ion more strongly / stronger ionic bonding

2. a) Write the equation, including state symbols, for the process corresponding to the enthalpy of solution of magnesium chloride.

MgCl2(s) → Mg2+(aq) + 2Cl- (aq)

2 c) Solubility is the measure of how much of a substance can be dissolved in water to make a saturated solution. A salt solution is saturated when an undissolved solid is in equilibrium with its aqueous ions. Use your answer to part (b) to deduce how the solubility of MgCl2 changes as the temperature is increased. Explain your answer.

Solubility will decrease as temperature increases. The enthalpy of solution is exothermic. So according to Le Chattier, the equilibrium will move to oppose the increase in temperature.

5. a) A 5.00 g sample of potassium chloride was added to 50.0 g of water initially at 20.0 °C. The mixture was stirred and as the potassium chloride dissolved, the temperature of the solution decreased. Describe the steps you would take to determine an accurate minimum temperature that is not influenced by heat from the surroundings.

Start a clock when KCl is added to water Record the temperature every subsequent minute for about 5 minutes Plot a graph of temperature vs time Extrapolate back to time of mixing = 0 and determine the temperature

ii) The theoretical enthalpy of lattice dissociation for silver chloride is +770 kJ mol-1. Explain why this value is less than the value for silver fluoride.

The chloride ion is larger than the fluoride So the attraction between Ag+ and Cl- is weaker

(b) In terms of electrostatic forces, suggest why the electron affinity of fluorine has a negative value.

There is an attraction between the protons in the nucleus and the added electrons and energy is released when an electron is gained.

(b) Calculate a value for the temperature when the reaction becomes feasible.

When ΔG = 0, ΔH = TΔS therefore T = ΔH / ΔS = -49 × 1000 / -180 = 272 (K)

5 (c) The enthalpy of solution of CaCl2 is −82.9 kJ mol−1. The enthalpies of hydration for calcium ions and chloride ions are −1650 and −364 kJ mol−1, respectively. Use these values to calculate a value for the lattice enthalpy of dissociation of calcium chloride.

ΔHsolution = ΔHlattice + ΔH(hydration of calcium ions) + 2 × ΔH(hydration of chloride ions) ΔHlattice = ΔHsolution - ΔH(hydration of calcium ions) -2 ×ΔH(hydration of chloride ions) ΔHlattice = -82-9 - (-1650 + 2 × -364) = +2295 (kJ mol-1)

4. a) Use this enthalpy change and data from the table to calculate a value for the free-energy change of the reaction at 250 °C.

ΔS = 238 + 189 - 214 - 3 × 131 = -180 J K-1 mol-1 ΔG = ΔH - TΔS = -49 - (523 x -180)/1000 = +45.1 kJ mol-1

3. (d) Use your answers to parts (a) and (b) to calculate a value for the free-energy change for this reaction at 50°C.

∆GΘ = ∆HΘ - T∆SΘ = -196 - 323 (-189/1000) = -134.9 kJ mol-1 Feasible because ∆G is negative

2 b) Use these data to calculate the standard enthalpy of solution of magnesium chloride. Enthalpy of lattice dissociation of MgCl2 = +2493 kJ mol-1 Enthalpy of hydration of magnesium ions = -1920 kJ mol-1 Enthalpy of hydration of chloride ions = -364 kJ mol-1

∆Hsoln MgCl2 = LE + ( ∆HhydMg2+) + 2( ∆HhydCl-) ∆Hsoln MgCl2 = 2493 - 1920 + (2 × -364) = -155 (kJ mol-1)

3. a) In the Contact Process sulfur dioxide reacts with oxygen to form sulfur trioxide as shown in the equation. 2SO2(g) + O2(g) ⇌ 2SO3(g) Use data from the table to calculate the standard enthalpy change for this reaction. b) calculate the standard entropy change. (c) State what the sign of the entropy change in your answer to part (b) indicates about the product of this reaction relative to the reactants.

∆HΘ = Σ∆ products - Σ∆ reactants (2 × -395) - (2 × -297) = -196 (kJ mol-1) ∆SΘ = ΣSΘ products - ΣSΘ reactants = (2 × 256) - 205 - (2 × 248) = -189 JK-1 mol-1 Negative sign = Causes an increase in order / a decrease in disorder

6 b) Use your Born−Haber cycle from part (a) and data from Table 1 to calculate a value for the electron affinity of chlorine.

−2EA(Cl) = 642 + 150 + 736 + 1450 + 242 − 2493 = 727 1 EA(Cl) = −364 (kJ mol−1 )