Experiment 7_II.doc

Experiment 7_II.doc - EXPERIMENT 7 Thermodynamics of...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
EXPERIMENT 7 Thermodynamics of Solution 55 E XPERIMENT 7 Thermodynamics of Solution 7.1 Purpose In experiment 7, values for enthalpy and entropy of solution of an inorganic salt will be determined, based on the temperature dependence of its solubility product K SP . 7.2 Background To understand why things dissolve at all, it is helpful to look at the solution formation process from a thermodynamic point of view. We therefore begin by considering a thermodynamic cycle that represents the formation of a solution from the isolated solute and solvent, as outlined in Figure 7-1. Figure 7-1 : Break-down of the solution process into individual steps The three distinct steps that we consider in solution formation are separation of solute molecules, separation of solvent molecules, and formation of solute-solvent interactions. Each step is associated with a change in enthalpy, Δ H 1 , Δ H 2 and Δ H 3 , respectively.
Background image of page 1

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

View Full DocumentRight Arrow Icon
EXPERIMENT 7 Thermodynamics of Solution 56 Δ H 1 and Δ H 2 are both positive because it requires energy to pull molecules or ions away from each other. That energy cost is due to the intermolecular forces present within any solute or solvent, such as electrostatic forces, dipole-dipole interactions, and hydrogen bonding. Each of these forces increases with decreasing distance. Therefore, it costs energy to pull molecules and ions away from each other. When the expanded form of the solvent and the solute are combined to form a solution, energy is released, causing Δ H 3 to be negative. This is due to the fact the solute and solvent now interact with each other through the various types of intermolecular forces. According to Hess’ Law of Constant Heat Summation we can express the energy enthalpy of solution Δ H sol as follows (equation 7.1): Δ H sol = Δ H 1 + Δ H 2 + Δ H 3 ( 7 . 1 ) In Figure 7-2, Hess’ Law is illustrated for an exothermic and for an endothermic solution process. Figure 7-2 : Illustration of Hess’ Law for a solution formation process. What determines the enthalpy of solution Δ H sol is, therefore, the difference between the energy required to separate the solvent and solute and energy released when the separated solvent and solute form a solution. To restate this in simpler terms, solutions will form when the energy of interaction between the solvent and solute is greater than the sum of the solvent-solvent and solute-solute interactions.
Background image of page 2
EXPERIMENT 7 Thermodynamics of Solution 57 In order to decide whether a solution process constitutes a spontaneous reaction, we have to consider the change in free energy, Δ G, since for a spontaneous process at constant temperature and pressure, Δ G < 0. The Gibbs equation describes how changes in free energy Δ G can be obtained from changes in enthalpy Δ H and entropy Δ S, equation 7.2. Δ
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/29/2008 for the course CHEM 118 taught by Professor Jacobsen during the Spring '08 term at Tulane.

Page1 / 8

Experiment 7_II.doc - EXPERIMENT 7 Thermodynamics of...

This preview shows document pages 1 - 4. Sign up to view the full document.

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