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Chapter 15: Problem 28

Outline the steps for calculating the concentrations of reacting species in an equilibrium reaction.

Short Answer

Expert verified
The concentrations of reacting species in an equilibrium reaction can be calculated by understanding the concept of chemical equilibrium, writing the correct equilibrium expressions using the law of mass action, and solving these expressions with the given values to obtain the concentrations.

Step by step solution

01

Understanding Chemical Equilibrium

In a chemical equilibrium, the rate of formation of products from reactants is equal to the rate of formation of reactants from products. This state is identified by constant, but not necessarily equal, concentrations of reactants and products.
02

Deriving the Equilibrium Expression

The equilibrium expression for a chemical reaction is derived using the law of mass action which states that the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients, is a constant at constant temperature. We represent the equilibrium constant with the symbol K.
03

Setting up the Equilibrium Expressions

Start by writing down the balanced chemical equation for the reaction. Consider the generic equation: aA + bB 鈬� cC + dD. The equilibrium expression for this reaction would be: \(K = [C]^c [D]^d / [A]^a [B]^b \) .
04

Calculating the Concentrations

To calculate the equilibrium concentrations of the reacting species, we substitute the known values (initial concentrations, K value) into the equilibrium expressions and solve for each concentration. If more than one concentration is unknown, we may need to set up and solve system of equations. The result will give the equilibrium concentrations of all reactants and products.

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Most popular questions from this chapter

Iodine is sparingly soluble in water but much more so in carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\). The equilibrium constant, also called the partition coefficient, for the distribution of \(\mathrm{I}_{2}\) between these two phases $$ \mathrm{I}_{2}(a q) \rightleftharpoons \mathrm{I}_{2}\left(\mathrm{CCl}_{4}\right) $$ is 83 at \(20^{\circ} \mathrm{C}\). (a) A student adds \(0.030 \mathrm{~L}\) of \(\mathrm{CCl}_{4}\) to \(0.200 \mathrm{~L}\) of an aqueous solution containing \(0.032 \mathrm{~g}\) \(\mathrm{I}_{2} .\) The mixture is shaken and the two phases are then allowed to separate. Calculate the fraction of \(\mathrm{I}_{2}\) remaining in the aqueous phase. (b) The student now repeats the extraction of \(I_{2}\) with another 0.030 \(\mathrm{L}\) of \(\mathrm{CCl}_{4}\). Calculate the fraction of the \(\mathrm{I}_{2}\) from the original solution that remains in the aqueous phase. (c) Compare the result in (b) with a single extraction using \(0.060 \mathrm{~L}\) of \(\mathrm{CCl}_{4}\). Comment on the difference.

Eggshells are composed mostly of calcium carbonate \(\left(\mathrm{CaCO}_{3}\right)\) formed by the reaction $$ \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftharpoons \mathrm{CaCO}_{3}(s) $$ The carbonate ions are supplied by carbon dioxide produced as a result of metabolism. Explain why eggshells are thinner in the summer, when the rate of chicken panting is greater. Suggest a remedy for this situation.

Consider the potential energy diagrams for two types of reactions \(\mathrm{A} \rightleftharpoons \mathrm{B}\). In each case, answer the following questions for the system at equilibrium. (a) How would a catalyst affect the forward and reverse rates of the reaction? (b) How would a catalyst affect the energies of the reactant and product? (c) How would an increase in temperature affect the equilibrium constant? (d) If the only effect of a catalyst is to lower the activation energies for the forward and reverse reactions, show that the equilibrium constant remains unchanged if a catalyst is added to the reacting mixture.

At \(25^{\circ} \mathrm{C}\), a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases are in equilibrium in a cylinder fitted with a movable piston. The concentrations are: \(\left[\mathrm{NO}_{2}\right]=0.0475 \mathrm{M}\) and \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.491 \mathrm{M}\). The volume of the gas mixture is halved by pushing down on the piston at constant temperature. Calculate the concentrations of the gases when equilibrium is reestablished. Will the color become darker or lighter after the change? [Hint: \(K_{\mathrm{c}}\) for the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is \(4.63 \times\) \(10^{-3} . \mathrm{N}_{2} \mathrm{O}_{4}(g)\) is colorless and \(\mathrm{NO}_{2}(g)\) has a brown color.]

When dissolved in water, glucose (corn sugar) and fructose (fruit sugar) exist in equilibrium as follows: fructose \(\rightleftharpoons\) glucose A chemist prepared a \(0.244 M\) fructose solution at \(25^{\circ} \mathrm{C}\). At equilibrium, it was found that its concentration had decreased to \(0.113 M\). (a) Calculate the equilibrium constant for the reaction. (b) At equilibrium, what percentage of fructose was converted to glucose?
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