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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 ⇌ 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

In the uncatalyzed reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ at \(100^{\circ} \mathrm{C}\) the pressures of the gases at equilibrium are \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.377 \mathrm{~atm}\) and \(P_{\mathrm{NO}_{2}}=1.56 \mathrm{~atm} .\) What would happen to these pressures if a catalyst were present?

Consider the reaction $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ At \(430^{\circ} \mathrm{C}\), an equilibrium mixture consists of 0.020 mole of \(\mathrm{O}_{2}, 0.040\) mole of \(\mathrm{NO},\) and 0.96 mole of \(\mathrm{NO}_{2} .\) Calculate \(K_{P}\) for the reaction, given that the total pressure is 0.20 atm.

The equilibrium constant \(K_{P}\) for the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ is found to be \(2 \times 10^{-42}\) at \(25^{\circ} \mathrm{C}\). (a) What is \(K_{\mathrm{c}}\) for the reaction at the same temperature? (b) The very small value of \(K_{P}\left(\right.\) and \(\left.K_{\mathrm{c}}\right)\) indicates that the reaction overwhelmingly favors the formation of water molecules. Explain why, despite this fact, a mixture of hydrogen and oxygen gases can be kept at room temperature without any change.

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.]

Consider this equilibrium process: $$ \begin{aligned} \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \\\ \Delta H^{\circ} &=92.5 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Predict the direction of the shift in equilibrium when (a) the temperature is raised, (b) more chlorine gas is added to the reaction mixture, \((\mathrm{c})\) some \(\mathrm{PCl}_{3}\) is removed from the mixture, (d) the pressure on the gases is increased, (e) a catalyst is added to the reaction mixture.
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