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Sunday, April 14, 2013

CHEMISTRY PRE-UNIVERSITY

Experiment 3

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Experiment 4
Topic: Reaction kinetics.
Purpose: To determine the effect of temperature on the reaction rate.
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Experiment 5 
Topic: Equilibrium and solubility
Purpose: To determine the solubility product, Ksp, of MX2 and enthalpy change of solution, ^Hsoln
General Rules with Ksp
In general, if the ion product is LESS THAN the Ksp, the solution is less than saturated
and there is no precipitate (ppt). If the ion product is EQUAL TO the Ksp, the solution is
saturated and there is a very fine equilibrium that is constantly changing between no
precipitate and a micro-precipitate. This equilibrium is so dynamic that one never
observes the ppt. If the ion product is GREATER THAN the Ksp, the solution is supersaturated and there is a readily observable ppt.
Calculating Ksp‘s from Solubility Data
In order to calculate the Ksp for an ionic compound you need the equation for the dissolving process so the equilibrium expression can be written.  You also need the concentrations of each ion expressed in terms of molarity, or moles per liter, or the means to obtain these values.
Example:  Calculate the solubility product constant for lead(II) chloride, if 50.0 mL of a saturated solution of lead(II) chloride was found to contain 0.2207 g of lead(II) chloride dissolved in it.
  • First, write the equation for the dissolving of lead(II) chloride and the equilibrium expression for the dissolving process.
PbCl2(s) –> Pb2+(aq) + 2 Cl-(aq)Ksp = [Pb2+][Cl-]2
  • Second, convert the amount of dissolved lead(II) chloride into moles per liter.
(0.2207 g PbCl2)(1/50.0 mL solution)(1000 mL/1 L)(1 mol PbCl2/278.1 g PbCl2) = 0.0159 M PbCl2
  • Third, create an “ICE” table.
PbCl2 (s)Pb2+(aq)Cl-(aq)
Initial ConcentrationAll solid00
Change in Concentration- 0.0159 M (dissolves)+ 0.0159 M+ 0.0318 M
Equilibrium ConcentrationLess solid0.0159 M0.0318 M
  • Fourth, substitute the equilibrium concentrations into the equilibrium expression and solve for Ksp.
Ksp = [0.0159][0.0318]2 = 1.61 x 10-5

Sample Table of Standard Enthalpy of Formation Values

This table provides a few sample values of the standard enthalpies of formation of various compounds:

 Compound ΔHfo
O2(g)0 kJ/mol
 C(graphite)  0 kJ/mol
 CO(g) -110.5 kJ/mol
 CO2(g) -393.5 kJ/mol
 H2(g) 0 kJ/mol
 H2O(g) -241.8 kJ/mol
 HF(g) -271.1 kJ/mol
 NO(g) 90.25 kJ/mol
NO2(g)33.18 kJ/mol
 N2O4(g) 9.16 kJ/mol
 SO2(g) -296.8 kJ/mol
 SO3(g) -395.7 kJ/mol


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Prepared By Cikgu Hanif
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