Solving Equilibrium Problems

Solving Equilibrium Problems-38
The only one who can teach you how to interpret, understand, and solve problems is yourself.

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Pressure measurements are ordinarily able to measure gonly 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.

Solution: 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.

Solution: As always, set up a table showing what you know (first two rows) and then expressing the equilibrium quantities: .

The term in the denominator of each mole fraction is the total number of moles of gas present at equilibrium: (0.200 – x) (3.00 x) x = 3.20 x.

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.

The biochemical formation of a disaccharide (double) sugar from two monosaccharides is exemplified by the reaction fructose glucose-6-phosphate → sucrose-6-phosphate (Sucrose is ordinary table sugar.) To what volume should a solution containing 0.050 mol of each monosaccharide be diluted in order to bring about 5% conversion to sucrose phosphate?

Solution: The initial and final numbers of moles are as follows: It often happens that two immiscible liquid phases are in contact, one of which contains a solute.

How will the solute tend to distribute itself between the two phases?

One’s first thought might be that some of the solute will migrate from one phase into the other until it is distributed equally between the two phases, since this would correspond to the maximum dispersion (randomness) of the solute.

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