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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] [Clâ‚‚] 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 * [Clâ‚‚]) / [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 [Clâ‚‚]_eq = (1.48 * 10^(-3)) M The equilibrium concentrations for all species are: [NOCl] = 2.59704 M [NO] = 2.96 * 10^(-3) M [Clâ‚‚] = 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 Clâ‚‚ 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 Clâ‚‚ 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 Clâ‚‚ 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 Clâ‚‚ 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 Clâ‚‚. 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

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$ \mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{H}_{2}(a q) $$ Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10}\), what are the equilibrium concentrations of each species if you start with a \(1.24 M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

At \(1100 \mathrm{~K}, K_{\mathrm{p}}=0.25\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ What is the value of \(K\) at this temperature?

Write expressions for \(K_{\mathrm{p}}\) for the following reactions. a. \(2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) b. \(\mathrm{CO}_{2}(g)+\mathrm{MgO}(s) \rightleftharpoons \mathrm{MgCO}_{3}(s)\) c. \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) d. \(4 \mathrm{KO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{KOH}(s)+3 \mathrm{O}_{2}(g)\)

At a particular temperature, \(K=3.75\) for the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g) $$ If all four gases had initial concentrations of \(0.800 M\), calculate the equilibrium concentrations of the gases.

At \(2200^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.050\) for the reaction $$ \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ What is the partial pressure of \(\mathrm{NO}\) in equilibrium with \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) that were placed in a flask at initial pressures of \(0.80\) and \(0.20\) atm, respectively?

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