Acid-base Chemistry and Water

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  • Topic: Acid-base chemistry, Acids, PH
  • Pages : 17 (5682 words )
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  • Published : October 28, 2010
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1  Measuring and calculating equilibrium constants
Clearly, if the concentrations or pressures of all the components of a reaction are known, then the value of K can be found by simple substitution. Observing individual concentrations or partial pressures directly may be not always be practical, however. If one of the components is colored, the extent to which it absorbs light of an appropriate wavelength may serve as an index of its concentration. Pressure measurements are ordinarily able to measure only the total pressure of a gaseous mixture, so if two or more gaseous products are present in the equilibrium mixture, the partial pressure of one may need to be inferred from that of the other, taking into account the stoichiometry of the reaction. Problem Example 1

In an experiment carried out by Taylor and Krist (J. Am. Chem. Soc. 1941: 1377), hydrogen iodide was found to be 22.3% dissociated at 730.8°K. Calculate Kc for 2 HI(g) → H2(g) + I2 .
No explicit molar concentrations are given, but we do know that for every n moles of HI, 0.223n moles of each product is formed and (1–0.233)n = 0.777n moles of HI remains. For simplicity, we assume that n=1 and that the reaction is carried out in a 1.00-L vessel, so that we can substitute the required concentration terms directly into the equilibrium expression. [pic]

Problem Example 2
Ordinary white phosphorus, P4, forms a vapor which dissociates into diatomic molecules at high temperatures:   P4(g)→ 2 P2(g) A sample of white phosphorus, when heated to 1000°C, formed a vapor having a total pressure of 0.20 atm and a density of 0.152 g L–1. Use this information to evaluate the equilibrium constant Kp for this reaction. Solution:  Before worrying about what the density of the gas mixture has to do with Kp , start out in the usual way by laying out the information required to express Kp in terms of an unknown x.  

| |P4 |2 P2 | | |initial moles: |1 |1 - x |Since K is independent of the number of moles, assume the | | | | |simplest case. | |moles at equilibrium: |1 - x |2x |x is the fraction of P4 that dissociates. | |eq. mole fractions: |[pic] |[pic] |The denominator is the total number of moles: | | | | |(1-x) + 2x = 1+x. | |eq. partial pressures: |[pic] |[pic] |Partial pressure is the mole fraction times the total | | | | |pressure. |

Expressing the equilibrium constant in terms of x gives
Now we need to find the dissociation fraction x of P4, and at this point we hope you remember those gas laws that you were told you would be needing later in the course! The density of a gas is directly proportional to its molecular weight, so you need to calculate the densities of pure P4 and pure P2 vapors under the conditions of the experiment. One of these densities will be greater than 0.152 gL–1 and the other will be smaller; all you need to do is to find where the measured density falls in between the two limits, and you will have the dissociation fraction. The molecular weight of phosphorus is 31.97, giving a molar mass of 127.9 g for P4. This mass must be divided by the volume to find the density; assuming ideal gas behavior, the volume of 127.9 g (1 mole) of P4 is given by RT/P, which works out to 522 L (remember to use the absolute temperature here.) The density of pure P4 vapor under the conditions of the experiment is then d = m/V =...
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