As we previously stated, lattice energy represents the union of a gaseous cation and a gaseous anion in the formation of an ionic solid.

Concept #1: Lattice Energy Application Intro

Concept #2: Lattice Energy Application

Example #1: Lattice Energy Application

Example #2: Lattice Energy Application

Use data from table , the figure and the figure to calculate the lattice energy of RbCl.

Potassium perchlorate (KClO4) has a lattice energy of -599 kJ/mol and a heat of hydration of -548 kJ/mol.Find the heat of solution for potassium perchlorate when 10.6 g of potassium perchlorate is dissolved with enough water to make 110.1 mL of solution.

Potassium perchlorate (KClO4) has a lattice energy of -599 kJ/mol and a heat of hydration of -548 kJ/mol.Determine the temperature change that occurs when 10.6 g of potassium perchlorate is dissolved with enough water to make 110.1 mL of solution. (Assume a heat capacity of 4.05 J/goC for the solution and a density of 1.05 g/mL.)

Based on data in Table 8.2 in the textbook, estimate (within 30) the lattice energy forLiBr

Based on data in Table 8.2 in the textbook, estimate (within 30) the lattice energy forCsBr

Based on data in Table 8.2 in the textbook, estimate (within 30) the lattice energy forCaCl2

The substances NaF and CaO are isoelectronic (have the same number of valence electrons).Predict the lattice energy of MgCl2.

The ionic compound CaO crystallizes with the same structure as sodium chloride.
From
the ionic radii given in the figure, calculate the potential energy of a single Ca-O ion pair that is just touching.
e = 1.60 10-19 C
= 8.99 109 J m /C2

The ionic compound CaO crystallizes with the same structure as sodium chloride.
Calculate the energy of a mole of such Ca-O ion pairs.

Lithium iodide has a lattice energy of −7.3 × 102kJ/mol and a heat of hydration of −793kJ/mol. Determine how much heat is evolved or absorbed when 40.0 g of lithium iodide completely dissolves in water.

Lithium iodide has a lattice energy of -7.3 x 102 kJ/mol and a heat of hydration of -793 kJ/mol. Find the heat of solution for lithium iodide and determine how much heat is evolved or absorbed when 15.0 g of lithium iodide completely dissolves in water.

Lithium iodide has a lattice energy of -7.3 x 10 2 kJ/mol and a heat of hydration of - 793 kJ/mol.a. Find the heat of solution for lithium iodide.b. Determine how much heat is evolved or absorbed when 40.0 g of lithium iodide completely dissolves in water.c. Is the heat evolved or absorbed?

Silver nitrate has a lattice energy of -820 kJ/mol and a heat of solution of 22.6 kJ/mol.Calculate the heat of hydration of silver nitrate?

Use the following information to calculate the ΔH° lattice of MgF2:Mg(s) → Mg(g) ΔH° =148 kJF2(g) → 2F(g) ΔH° =159 kJMg(g) → Mg+ (g)+e- ΔH° =738 kJMg+(g) → Mg2+ (g)+e- ΔH° =1450 kJF(g)+e- → F-(g) ΔH° =-328 kJMg(s) +F2(g) → MgF2 (s) ΔH° =-1123 kJ_______ kJ

Which of the following values of ΔH lattice and ΔHhydration would describe a salt which dissolves endothermically?A. ΔHlattice = +370 kJ/mol and ΔH hydration = -360kJ/molB. ΔHlattice = +390kJ/mol and ΔH hydration= -410 kJ/molC. ΔHlattice= +100 kJ/mol and ΔH hydration = -120 kJ/molD. ΔHlattice= +470 kJ/mol and ΔH hydration = -510 kJ/mol

Silver nitrate has a lattice energy of -820 kJ/mol and a heat of solution of + 22.6 kJ/mol. Calculate the heat of hydration for silver nitrate.

In a coffee-cup calorimeter, 1.60 g NH 4NO3 was mixed with 75.0 g water at an initial temperature 25.00˚C. After dissolution of the salt, the final temperature of the calorimeter contents was 23.34˚C. If the enthalpy of hydration for NH4NO3 is 2630. kJ/mol, calculate the lattice energy of NH4NO3.

Use the data to calculate the heats of hydration of lithium chloride and sodium chloride.

The lattice energy* of NaI is -686 kJ/mol, and the enthalpy of hydration is -694 kJ/mol. Calculate the enthalpy of solution per mole of solid NaI. Describe the process to which this enthalpy change applies.*Lattice energy was defined in Chapter 3 as the energy change for the process M+(g) + X-(g) → MX(s).

Lithium iodide has a lattice energy of -7.3 x 102 kJ/mol and a heat of hydration of -793 kJ/mol. Find the heat of solution for lithium iodide.

a. Use the following data to calculate the enthalpy of hydration for calcium chloride and calcium iodide.

Lithium iodide has a lattice energy of - 7.3 x 102 kJ/mol and a heat of hydration of -793 kJ/mol. Determine how much heat is evolved or absorbed when 30.0 g of lithium iodide completely dissolves in water.

(a) Use the following data to calculate the combined heat of hydration for the ions in potassium bromate (KBrO3): ΔHlattice = 745 kJ/mol ΔHsoln = 41.1 kJ/mol(b) Which ion contributes more to the answer for part (a)? Why?

Besides being used in black-and-white film, silver nitrate (AgNO3) is used similarly in forensic science. The NaCl left behind in the sweat of a fingerprint is treated with AgNO3 solution to form AgCl. This precipitate is developed to show the blackand-white fingerprint pattern. Given that ΔHlattice = 822 kJ/mol and ΔHhydr = −799 kJ/mol for AgNO3, calculate its ΔHsoln.

Sodium hydroxide (NaOH) has a lattice energy of -887 kJ/mol and a heat of hydration of -932 kJ/mol.How much solution could be heated to boiling by the heat evolved by the dissolution of 24.5 g of NaOH? (For the solution, assume a heat capacity of 4.0 J/goC, an initial temperature of 25.0 oC, a boiling point of 100.0 oC, and a density of 1.05 g/mL.)

Use the following to calculate ΔH°lattice of NaCl:Na(s) ⟶Na(g) ΔH° = 109 kJCl2(g) ⟶2Cl(g) ΔH° = 243 kJNa(g) ⟶Na+(g) + e− ΔH° = 496 kJCl(g) + e− ⟶Cl−(g) ΔH° = −349 kJNa(s) + 1/2 Cl2(g) ⟶NaCl(s) ΔH° = −411 kJCompared with the lattice energy of LiF (1050 kJ/mol), is the magnitude of the value for NaCl what you expected? Explain.

The lattice energy of NaCl is 2786 kJ/mol, and the enthalpy of hydration of 1 mole of gaseous Na+ and 1 mole of gaseous Cl – ions is 2783 kJ/mol. Calculate the enthalpy of solution per mole of solid NaCl.

Use the following to calculate ΔH°lattice of MgF2:Mg(s) ⟶Mg(g) ΔH° = 148 kJF2(g) ⟶2F(g) ΔH° = 159 kJMg(g) ⟶Mg+(g) + e− ΔH° = 738 kJMg+(g) ⟶Mg2+(g) + e− ΔH° = 1450 kJF(g) + e− ⟶F−(g) ΔH° = −328 kJMg(s) + F2(g) ⟶MgF2(s) ΔH° = −1123 kJCompared with the lattice energy of LiF (1050 kJ/mol) or the lattice energy calculated for NaCl below, does the relative magnitude of the value for MgF2 surprise you? Explain.Na(s) ⟶Na(g) ΔH° = 109 kJCl2(g) ⟶2Cl(g) ΔH° = 243 kJNa(g) ⟶Na+(g) + e− ΔH° = 496 kJCl(g) + e− ⟶Cl−(g) ΔH° = −349 kJNa(s) + 1/2 Cl2(g) ⟶NaCl(s) ΔH° = −411 kJ

You may want to reference (Pages 382 - 416) Chapter 9 while completing this problem.NaCl has a lattice energy -787 kJ/mol. Consider a hypothetical salt XY. X2+ has the same radius of Na+ and Y2– has the same radius as Cl–. Estimate the lattice energy of XY.

Potassium nitrate has a lattice energy of -163.8 kcal/mol and a heat of hydration of -155.5 kcal/mol. How much potassium nitrate has to dissolve in water to absorb 111 kJ of heat?

Silver nitrate has a lattice energy of 820.kJ/mol and a heat of solution of +22.6kJ/mol. Calculate the heat of hydration for silver nitrate.

Lithium iodide has a lattice energy of −7.3 x 102 kJ/mol and a heat of hydration of −793kJ/mol. Find the heat of solution for lithium iodide.

(a) Use the following data to calculate the combined heat of hydration for the ions in sodium acetate (NaC2H3O2): ΔHlattice = 763 kJ/mol ΔHsoln = 17.3 kJ/mol(b) Which ion contributes more to the answer for part (a)? Why?

Use the data below to calculate the heat of hydration of lithium chloride.Compound Lattice energy (kJ/mol) ΔHsoln (kJ/mol)LiCl-834-37.0NaCl-769+3.88