13.4 Equilibrium Calculations

13.4 Equilibrium Calculations

• Heavy equipment is required to handle high temperatures and pressures.
• The design of an ammonia plant is outlined in this schematic.

• The equilibrium constant is equal to the value of the reaction quotient.

• We want to extend our understanding of which direction a reaction will shift to reach equilibrium by using quantitative calculations.
• We look at the ways in which the concentrations of products and reactants change as a reaction approaches equilibrium.
• This approach to equilibrium calculations will be explored in this section.

• As a reaction system approaches equilibrium, there can be changes in concentrations and pressures of reactants and products.

• We can use the coefficients in the balanced chemical equation to relate the changes to each other.
• The decomposition of ammonia is an example.

• The change is positive because the concentration of N2 increases.

• The change in the H2 concentration, D[H2], is positive.
• The change in the concentration of H2 is three times the change in the concentration of N2 because for each mole of N2 produced, 3 moles of H2 are produced.

• The equation shows that 2 moles of NH3 must be formed for each mole of N2 to be dissolved.

• The changes on one side of the arrows are the same sign as the changes on the other side.

• The coefficients in the D terms are the same as those in the balanced equation.

• The simplest way to find the coefficients for the concentration changes in a reaction is to use the balanced chemical equation.
• When the concentration increases, the sign is positive, while when it decreases, it is negative.

• Change the concentrations for each reaction.

• The equilibrium constant can be calculated if concentrations of reactants and products are known.

• The remaining concentration can be calculated if the equilibrium constant and all of the equilibrium concentrations are known.

• Concentrations at equilibrium can be calculated if the equilibrium constant and a set of concentrations of reactants and products are known.

• Each case will be solved in a sequence.

• The following example shows how to use a combination of initial concentrations and equilibrium concentrations to determine an equilibrium constant.

• The equilibrium reaction is the subject of a chart.
• Underneath the reaction the initial concentrations of the reactants and products are listed.
• As the system shifts toward equilibrium, the next row of data is the change that occurs.
• Once equilibrium has been reached, the concentrations are in the last row.

• The triiodide ion is produced by the reversibly reacting iodine molecule with the ion.

• As the system goes to equilibrium, we will calculate the changes in concentration.

• The equilibrium constant is determined after we determine the equilibrium concentrations.

• The table can now be filled with the concentrations at equilibrium.

• The equilibrium constant is now calculated.

• We can calculate the missing concentration if we know the equilibrium constant for a reaction and the concentrations of all reactants and products.

• Nitrogen oxides can be produced by the reaction of nitrogen and oxygen.

• In air, [N2] is 0.036mol/L and [O2] is 0.0089mol/L.

• All of the equilibrium concentrations are given to us.
• The missing equilibrium concentration can be solved by rearranging the equation for the equilibrium constant.

• The NO is 3.6 x 10-4 mol/L at equilibrium.

• We can check our answer by substituting equilibrium concentrations into the expression for the reaction quotient to see if it is equal to the equilibrium constant.

• The equilibrium constant within the error associated with the significant figures in the problem is given by our calculated value.

• The procedure can be summarized in four steps.

• The direction the reaction proceeds should be determined.

• A balanced chemical equation is needed for the reaction.

• Write the equilibrium concentrations in terms of the relative changes needed to reach equilibrium.

• The reaction to reach equilibrium requires changes in the initial concentrations.

• Define missing equilibrium concentrations in terms of the initial concentrations and the changes in concentration.

• Check the equilibrium concentrations by substituting them into the equilibrium expression to see if they give the equilibrium constant.

• Sometimes a particular step may be more complex in some problems and less complex in others.
• Every calculation of equilibrium concentrations from a set of initial concentrations will involve steps.

• Sometimes it is convenient to set up an ICE table in order to solve equilibrium problems that involve changes in concentration.

• The stepwise process was described earlier.

• Determine the direction of the reaction.

• The reaction will proceed to the right.

• There are two different solutions to a quadratic equation, one that is physically possible and one that is physically impossible.
• The second solution is impossible because we know the change must be a positive number, otherwise we would end up with negative values for concentrations of the products.

• The equilibrium concentrations are checked.

• CH3CO2H reacts with C2H5OH to form water and CH3CO2C2H5.

• There is an equilibrium constant for this reaction.

• There are 1.00 moles of H2 and 1.00 moles of I2 in a 1.00-L flask.
• Under the given conditions, the equilibrium constant is 50.5 for the reaction of hydrogen and iodine to form hydrogen iodide.

• Sometimes it is possible to use chemical insight to find solutions to equilibrium problems without actually solving the equation.

• The weak acid is 0.150 M HA.

• The easiest way to determine the equilibrium concentrations is to start with only reactants.
• This could be called the "all reactant" starting point.

• A less obvious way to solve the problem would be to assume all the HA ionizes first.
• This could be called the "all product" starting point.
• The ICE table is used to represent the concentration of HA at equilibrium.

• To minimize rounding problems, keep a few extra significant figures.

• Small is defined as an error of less than 5%.
• Two examples demonstrate this.

• The concentration of each product is represented by the ICE table.

• Chemical intuition can provide a simpler solution.

• The same result was given to two significant figures when the exact (quadratic) equation and using approximations were solved.
• The approximate solution is valid if the error is less than 5%.

• The approximate solution is a valid solution.

• The 5% rule requires you to pick the smallest concentration.

• The assumptions are valid because this is less than 5%.

• When the amounts of reactants and products don't change, a reaction is at equilibrium.
• The rate of formation of products by the forward reaction is equal to the rate at which the products re-form reactants by the reverse reaction.

• If a reactant or product is a pure solid, a pure liquid, or the solvent in a dilute solution, the concentration of this component does not appear in the expression for the equilibrium constant.
• The values of the reactants and products are constant.

• All components are in the same phase of the equilibrium.
• The components are in two or more phases in a heterogeneous equilibrium.
• If we compare the reaction quotient with the equilibrium constant, we can determine if a reaction is at equilibrium.

• If the number of moles of gas is different on the reactant and product sides of the reaction, equilibrium can be disturbed.
• Le Chatelier's principle states that the system will respond in a way that counteracts the disturbances.
• Adding a catalyst affects the rates of the reactions but does not alter the equilibrium, and changing pressure or volume will not disturb systems with no gases or equal numbers of moles of gas on the reactant and product side.

• The ratios of the rate of change in concentrations of a reaction are the same as the coefficients in the balanced chemical equation.
• When the concentration increases and decreases, the sign of the coefficient of X is positive.
• We were able to approach three basic types of equilibrium problems.
• We can solve for the equilibrium constant, when we have the equilibrium constant and some of the concentrations involved, and for the missing concentrations, when we have the equilibrium constant and the initial concentrations.

• The statement that all chlorides aresoluble is one of the solubility rules.

• The statement: Carbonates, phosphates, borates, and arsenates--except those of the ammonium ion and the alkali metals--are insoluble.

• Benzene is used as an enhancer in gasoline.

• The ion K+ and I - 3 are in KI3.

• The yield of the reaction must essentially be 100% for a titration to be effective.

• The product of a precipitation reaction must be insoluble for it to be useful.

• The initial concentrations of reactants and products are given for each system.

• The initial concentrations of reactants and products are given for each system.

• The following system is at an equilibrium.

• Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and hydrogen at high temperature and pressure.

• Water gas, a mixture of H2 and CO, is an important industrial fuel produced by the reaction of steam with red hot coke.

• Reduction of iron(III) oxide with hydrogen gas can be used to make pure iron metal.

• Suggest two ways in which the equilibrium concentration of Ag+ can be reduced in a solution of Na+, Cl-, Ag+, and NO - 3.

• A solution of silver ion and sulfate ion is added to a solution of solid silver sulfate.

• There are two isomers of the amino acid alanine.
• The solution of a-alanine will freeze at the lowest temperature when the two compounds are dissolved in the same amount of solvent.

• Hydrogen is prepared commercially by the reaction of methane and water.

• A sample of PCl5 is put into a heated vessel.

• The equilibrium constant should be calculated for the reaction.

• The sample was heated in a closed container.

• The vapor pressure of water is 0.196 atm.

• Change the concentrations or pressure for each reaction.

• Change the concentrations or pressure for each reaction.

• Reducing the oxide of cobalt with carbon monoxide can be used to make the metal.

• Carbon reacts with water.

• The change in concentration of N2O4 is small enough to be ignored.

• The change is small enough to be neglected.

• The change in concentration of CO2 is small enough to be ignored.

• The change is small enough to be neglected.

• The change in pressure of H2S is small enough to be ignored.

• The change is small enough to be neglected.

• The HCl come to equilibrium.

• The following equilibrium partial pressures are measured in a 3.0-L vessel.

• HbO2 is partially regulated by the concentration of H3O+ and dissolved CO2 in the blood.

• The first-order rate equation for the disappearance of sucrose is followed by the hydrolysis of the sugar to the sugars.

• The constant for the reaction is 1.36 x 105.
• The effective concentration of a solvent is 1.

• The density of trifluoroacetic acid was found to be 2.784 g/L.

• The OpenStax book is free and can be found at http://cnx.org/content/col11760/1.9.
• The total number of solute particles is related to osmotic pressure.