Stoichiometry Lab Report
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 …show more content…
partial pressures: |[pic] |[pic] |[pic] |
The partial pressures in the bottom row were found by multiplying the mole fraction of each gas by the total pressure: Pi = XiPt. 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.
Substituting the equilibrium partial pressures into the equilibrium expression, we have
[pic]
whose polynomial form is 4.60 x2 + 13.80 x – 2.304 = …show more content…
This is because as the solution becomes more dilute, the product [H3O+][Ac–] decreases more rapidly than does the [HAc] term. At the same time the concentration of H2O becomes greater, but because it is so large to start with (about 55.5M), any effect this might have is negligible, which is why no [H2O] term appears in the equilibrium expression.
For a reaction such as CH3COOH(l) + C2H5OH(l) → CH3COOC2H5(l) + H2O(l) (in which the water concentration does change), dilution will have no effect on the equilibrium; the situation is analogous to the way the pressure dependence of a gas-phase reaction depends on the number of moles of gaseous components on either side of the equation.
Problem Example 5
The biochemical formation of a disaccharide (double) sugar from two monosaccharides is exemplified by the reaction fructose + glucose-6-phosphate →
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 …show more content…
partial pressures: |[pic] |[pic] |[pic] |
The partial pressures in the bottom row were found by multiplying the mole fraction of each gas by the total pressure: Pi = XiPt. 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.
Substituting the equilibrium partial pressures into the equilibrium expression, we have
[pic]
whose polynomial form is 4.60 x2 + 13.80 x – 2.304 = …show more content…
This is because as the solution becomes more dilute, the product [H3O+][Ac–] decreases more rapidly than does the [HAc] term. At the same time the concentration of H2O becomes greater, but because it is so large to start with (about 55.5M), any effect this might have is negligible, which is why no [H2O] term appears in the equilibrium expression.
For a reaction such as CH3COOH(l) + C2H5OH(l) → CH3COOC2H5(l) + H2O(l) (in which the water concentration does change), dilution will have no effect on the equilibrium; the situation is analogous to the way the pressure dependence of a gas-phase reaction depends on the number of moles of gaseous components on either side of the equation.
Problem Example 5
The biochemical formation of a disaccharide (double) sugar from two monosaccharides is exemplified by the reaction fructose + glucose-6-phosphate →