In a given experiment, \(5.2\) moles of pure NOCl was placed in an otherwise empty \(2.0-\mathrm{L}\) container. Equilibrium was established by the following reaction: $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \quad K=1.6 \times 10^{-5} $$ a. Using numerical values for the concentrations in the Initial row and expressions containing the variable \(x\) in both the Change and Equilibrium rows, complete the following table summarizing what happens as this reaction reaches equilibrium. Let \(x=\) the concentration of \(\mathrm{Cl}_{2}\) that is present at equilibrium. b. Calculate the equilibrium concentrations for all species.

Short Answer

Expert verified
The equilibrium concentrations for all species are: \[ [\mathrm{NOCl}] = 2.59704\,\mathrm{M} \] \[ [\mathrm{NO}] = 2.96 \times 10^{-3}\,\mathrm{M} \] \[ [\mathrm{Cl}_2] = 1.48 \times 10^{-3}\,\mathrm{M} \]

Step by step solution

01

Calculate the initial concentration of NOCl

To calculate the initial concentration of NOCl, we can use the following formula: Initial Concentration (M)= moles of NOCl / Volume of the container (in Liters) NOCl_initial = (5.2 moles) / (2.0 L) = 2.6 M
02

Set up the table and expressions for Initial, Change, and Equilibrium concentrations of all species

We can use the ICE (Initial, Change, Equilibrium) table to represent these concentrations: [NOCl] [NO] [Cl₂] Initial 2.6 M 0 M 0 M Change -2x +2x +x Equilibrium 2.6-2x 2x x Note: All concentrations are in moles per liter (M).
03

Calculate the value of x using the given value of K

Using the given value of K = 1.6 * 10^(-5), we can write the equilibrium expression as: K = ([NO]^2 * [Cl₂]) / [NOCl]^2 Substitute the expressions from the Equilibrium row into the equilibrium expression: (1.6 * 10^(-5)) = [(2x)^2 * (x)] / (2.6-2x)^2
04

Solve the equation for x

We will now solve the equation (1.6 * 10^(-5)) = [(2x)^2 * (x)] / (2.6-2x)^2 for x: (1.6 * 10^(-5)) * (2.6 - 2x)^2 = (4x^2) * (x) x = 1.48 * 10^(-3)
05

Calculate the equilibrium concentrations for all species

Now we'll substitute the value of x back into the expressions from the Equilibrium row and calculate the equilibrium concentrations for all species: [NOCl]_eq = 2.6 - 2(1.48 * 10^(-3)) = 2.59704 M [NO]_eq = 2(1.48 * 10^(-3)) = 2.96 * 10^(-3) M [Cl₂]_eq = (1.48 * 10^(-3)) M The equilibrium concentrations for all species are: [NOCl] = 2.59704 M [NO] = 2.96 * 10^(-3) M [Cl₂] = 1.48 * 10^(-3) M

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
The equilibrium constant, denoted as K, is a vital figure in chemistry as it quantifies the extent of a chemical reaction at equilibrium. For the reaction illustrated as
\(2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{Cl}_{2}(g)\),
the equilibrium constant expression is constructed by raising the concentrations of the products to their stoichiometric coefficients and dividing by the concentration of the reactants similarly raised. Symbolically, this is represented as:
\[K = \frac{[\mathrm{NO}]^2 \times [\mathrm{Cl}_{2}]}{[\mathrm{NOCl}]^2}\].
Here, the square brackets denote the concentration of the gaseous species at equilibrium, and the raised powers correspond to their respective coefficients in the balanced equation. In our problem, K is given as \(1.6 \times 10^{-5}\), signifying the reaction favors the reactants at equilibrium due to its small value.
Initial Concentration
Initial concentration refers to the amount of reactant or product present in a reaction mixture before any reaction has taken place. In our example, the initial concentration of NOCl is calculated by dividing the initial number of moles by the volume of the container. The calculation is simple:
\[\text{Initial Concentration} = \frac{\text{moles of NOCl}}{\text{Volume of the container}}\].
With \(5.2\) moles of NOCl in a \(2.0-\mathrm{L}\) container, the concentration starts at \(2.6\) M. Initially, NO and Cl₂ are not present hence their concentrations are \(0\) M.
ICE Table
An ICE table is an organized way to apply stoichiometry to track changes in concentration over the course of a reaction until it reaches equilibrium. ICE stands for Initial, Change, and Equilibrium. It starts with the Initial concentrations, then considers the Changes that occur as the system reaches equilibrium, and finally, the Equilibrium concentrations. In our case, the ICE table is set up with the initial concentrations of NOCl, changes (-2x for NOCl and +2x for NO, +x for Cl₂ due to the stoichiometry of the reaction), and expressions using x for the equilibrium concentrations. This provides the necessary framework to solve for x, where x represents the increase in concentration of Cl₂ at equilibrium. Once x is determined through the equilibrium constant expression, it can be plugged back into the equilibrium expressions to find the exact concentrations of all substances at equilibrium.
Equilibrium Concentration
Equilibrium concentration is the concentration of each reactant and product in a chemical reaction mixture at the state of equilibrium. By using the ICE table and the value of x determined from the equilibrium constant, one can calculate these concentrations. The equilibrium concentration of NOCl is obtained by subtracting 2x from its initial concentration, while that of NO and Cl₂ is obtained by multiplying their change by x. As shown in the solution steps, the final equilibrium concentrations are \(2.59704\) M for NOCl, \(2.96 \times 10^{-3}\) M for NO, and \(1.48 \times 10^{-3}\) M for Cl₂. Accurate calculation of these concentrations is critical for predicting how a system responds to changes in conditions and for applications in chemical engineering and research.

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Most popular questions from this chapter

Consider the following statements: "Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)\), for which at equilibrium \([\mathrm{A}]=2 M\) \([\mathrm{B}]=1 M\), and \([\mathrm{C}]=4 M .\) To a \(1-\mathrm{L}\) container of the system at equilibrium, you add 3 moles of \(\mathrm{B}\). A possible equilibrium condition is \([\mathrm{A}]=1 M,[\mathrm{~B}]=3 M\), and \([\mathrm{C}]=6 M\) because in both cases \(K=2 . "\) Indicate everything that is correct in these statements and everything that is incorrect. Correct the incorrect statements, and explain.

At a particular temperature, \(K=1.00 \times 10^{2}\) for the reaction $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ In an experiment, \(1.00\) mole of \(\mathrm{H}_{2}, 1.00 \mathrm{~mole}\) of \(\mathrm{I}_{2}\), and \(1.00\) mole of HI are introduced into a \(1.00-\mathrm{L}\) container. Calculate the concentrations of all species when equilibrium is reached.

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ If \(2.0\) moles of \(\mathrm{NO}\) and \(1.0\) mole of \(\mathrm{Cl}_{2}\) are placed into a \(1.0-\mathrm{L}\) flask, calculate the equilibrium concentrations of all species.

Consider the following reaction at a certain temperature: $$ 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) $$ An equilibrium mixture contains \(1.0\) mole of \(\mathrm{Fe}, 1.0 \times 10^{-3}\) mole of \(\mathrm{O}_{2}\), and \(2.0\) moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) all in a \(2.0-\mathrm{L}\) container. Calculate the value of \(K\) for this reaction.

Consider the following exothermic reaction at equilibrium: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ Predict how the following changes affect the number of moles of each component of the system after equilibrium is reestablished by completing the table below. Complete the table with the terms increase, decrease, or no change.

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