# Equilibrium Exp

**Topics:**Chemical equilibrium, Concentration, Equilibrium constant

**Pages:**16 (3927 words)

**Published:**December 17, 2012

In your text (Chang, 6th Ed) : Ch. 15 Chemical Equilibrium, esp. Section 15.3

Purpose: The Law of Mass Action will be examined via a series of samples using the same reaction, but different stating concentrations. The equilibrium constant, K, for each reaction will be calculated, demonstrating that K for a given reaction at a fixed temperature is a constant, independent of starting concentrations.

Background: For a general reaction aA + bB ↔ cC + dD, the Law of Mass Action (formulated by Guldberg and Waage, 1864) in terms of concentration is:

[pic] = K (Eqn. 1)

For gases, K can also be stated in terms of partial pressures, and hence is labeled as Kp: [pic] = Kp (Eqn. 2)

In either case, the terms must be referenced to standard states, 1 M or 1 atm, so the numbers in the equations are unitless, and the K’s are unitless. For a given reaction, different initial concentrations will yield differing final concentrations, but K, the ratio of products over reactants, all terms raised to their stoichiometric coefficients, is a constant at a fixed temperature. (K is a function of temperature.)

In this experiment, five samples of different initial concentrations will be prepared for the reaction:

Fe+3(aq) + SCN-(aq) ↔ FeSCN+2(aq) (Eqn. 3)

Equilibrium calculations are simplified by using an ICE (Initial-Change-Equilibrium) table. Since this reaction is run in aqueous solution, the K calculated will be Kc (K in terms of concentration), and all concentrations must be expressed in terms of M, or moles/liter, then reference to the standard state of 1M.

To calculate the equilibrium concentrations and K, the ICE table is:

Fe+3(aq) + SCN-(aq) ↔ FeSCN+2(aq)

ICoCo’ 0

C-x-x +x

ECo - xCo’- x x

where

Co = initial concentration of Fe+3(aq) {Co mol/L ( 1 mol/L = Co} Co’ = initial concentration of SCN-(aq) {Co’ mol/L ( 1 mol/L = Co( } x = change in concentration{ x mol/L ( 1 mol/L = x } in Fe+3, SCN- (loss) and FeSCN+2 (gain), since it is a 1:1:1 reaction

_ [FeSCN+2] =_ [x] = K (Eqn. 4) [Fe+3] [SCN-] [Co – x] [Co’- x]

The sample will be made by mixing known volumes of stock solutions, and adding enough water to bring the total volume up to 12.00 ml. (The solutions are all dilute, in the same solvent, so volumes can be considered additive.) The initial concentrations must be calculated at the instant of mixing. The dilution equation may be used

M1 V1 = M2 V2(Eqn. 5)

to calculate the initial concentrations in the mixtures.

Example:

Note: This example reaction is run at an elevated temperature, so the K and the Beer’s Law data will not be the same as what you will measure at ambient temperature.

A sample is mixed following the dilution pattern of:

Amount added to mixture

Stock solution of Fe+3(aq) = 2.00 x 10-3 M 5.00 ml

Stock solution of SCN-(aq) = 2.00 x 10-3 M 3.00 ml

Water 4.00 ml

Total volume: 12.00 ml

Calculation of initial concentrations:

M1 V1 = M2 V2

Fe+3(2.00 x 10-3 M)( 5.00 ml) = (M2 )(12.00 ml)

M2 = 8.33 x 10-3 M

8.33 x 10-3 M( 1 mol/L = 8.33 x 10-3 = Co

SCN-(2.00 x 10-3 M)( 3.00 ml) = (M2 )(12.00 ml)

M2 = 5.00 x 10-4 M

5.00 x 10-4 M( 1 mol/L = 5.00 x 10-4 = Co’

The last piece of information needed is x, the FeSCN+2(aq) final concentration. “x” is also the change in concentration of both Fe+3 and SCN-, since this is a 1:1:1 reaction. For every mole of FeSCN+2 made, Fe+3 and SCN- lose a mole. Since the volume is constant, the same statement can be made in terms of molarity: for every mole/liter of FeSCN+2 made, Fe+3 and SCN- lose a mole/liter.

“x” will be...

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