At a particular temperature, \(K=4.0 \times 10^{-7}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ In an experiment, \(1.0\) mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is placed in a \(10.0-\mathrm{L}\) vessel. Calculate the concentrations of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) when this reaction reaches equilibrium.

Short Answer

Expert verified
The equilibrium concentrations of N₂O₄ and NO₂ are approximately 0.0988 M and 2.4 × 10⁻³ M, respectively.

Step by step solution

01

Calculate the initial concentration of N₂O₄ (g)

To calculate the initial concentration of N₂O₄ (g), use the formula: $$ Concentration = \frac{n}{V} $$ where: n = number of moles V = volume in liters In our case, n = \(1.0\) mole and V = \(10.0\) L. Plug in the numbers to calculate the initial concentration: $$ [\mathrm{N}_{2} \mathrm{O}_{4}]_{initial}=\frac{1.0\text{ mole}}{10.0\text{ L}}=0.1\text{ M} $$
02

Set up the ICE table

The ICE table helps to organize the initial, change, and equilibrium concentrations of the reactants and products. In this case, the table will look like the following: $$ \begin{array}{ccccccc} &\boldsymbol{\mathrm{N}_{2} \mathrm{O}_{4}(g)} & & 2\boldsymbol{\mathrm{NO}_{2}(g)} \\ \textbf{Initial}& 0.1 & & 0\\ \textbf{Change} & -x & & +2x\\ \textbf{Equilibrium} & 0.1-x& &2x \end{array} $$ where x represents the change in concentration of the reactants and products as equilibrium is reached.
03

Write the expression for the equilibrium constant K

We can use the expression for the equilibrium constant K for the given reaction: $$ K = \frac{[\mathrm{NO}_{2}]^{2}}{[\mathrm{N}_{2} \mathrm{O}_{4}]} $$ At equilibrium, the concentrations will be as follows: $$ K=\frac{(2x)^2}{0.1-x} $$
04

Solve for x

Plug in the given value of K: $$ 4.0 \times 10^{-7} = \frac{(2x)^2}{0.1-x} $$ Now, solve for x: $$ x \approx 1.2 \times 10^{-3} $$
05

Determine the equilibrium concentrations

Use the values of x to calculate the equilibrium concentrations of N₂O₄ and NO₂: $$ [\mathrm{N}_{2} \mathrm{O}_{4}]_{equilibrium} = 0.1-1.2 \times 10^{-3} \approx 0.0988 \mathrm{\,M} $$ $$ [\mathrm{NO}_{2}]_{equilibrium} = 2x = 2 \times (1.2 \times 10^{-3}) \approx 2.4 \times 10^{-3} \mathrm{\,M} $$ At equilibrium, the concentrations of N₂O₄ and NO₂ are approximately 0.0988 M and 2.4 × 10⁻³ M, respectively.

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Key Concepts

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

ICE Table
Understanding chemical equilibrium can sometimes be a complex task, but with an ICE table (Initial, Change, Equilibrium), it becomes a lot more manageable. The ICE table is a systematic way to organize the changes in concentration of reactants and products during a reaction as it approaches equilibrium.

Let's walk through an example. Suppose we start with 1.0 mole of N₂O₄ in a 10.0 L vessel. Initially, we have no NO₂, so its concentration is 0. We represent the initial concentration of N₂O₄ as 0.1 M and NO₂ as 0 M. As the reaction proceeds towards equilibrium, N₂O₄ decomposes into NO₂, and we define this change in concentration as 'x' and '2x', respectively, in line with the stoichiometry of the reaction.

Therefore, at equilibrium, the concentration of N₂O₄ will be '0.1 - x' M and for NO₂, it will be '2x' M. Using an ICE table efficiently categorizes these stages, providing clear visualization of the process, which greatly simplifies the computation of equilibrium concentrations.
Equilibrium Constant
A cornerstone of chemical equilibrium is the equilibrium constant, denoted as K. It's a number that expresses the ratio of the concentration of the products to the reactants at equilibrium, each raised to the power of their stoichiometric coefficients.

For our N₂O₄ and NO₂ reaction, the equilibrium constant, K, would be formulated as follows:
\[ K = \frac{[\mathrm{NO}_{2}]^{2}}{[\mathrm{N}_{2} \mathrm{O}_{4}]} \]
In the real world of chemistry, a large K (much greater than 1) suggests that at equilibrium, mostly products are present, indicating the forward reaction is favored. Conversely, a small K (much less than 1) means that reactants predominate, and the reverse reaction is favored. For our reaction, K is given to be 4.0 x 10⁻⁷, indicating the equilibrium heavily favors the reactants, N₂O₄.
Reaction Quotient
Before a system reaches equilibrium, we can predict the direction in which the reaction will proceed using the reaction quotient (Q). Like the equilibrium constant, it represents a ratio of product and reactant concentrations at any point in time before the system reaches equilibrium, taking into account their stoichiometric coefficients.

The reaction quotient is calculated using the formula:
\[ Q = \frac{[\mathrm{NO}_{2}]^{2}}{[\mathrm{N}_{2} \mathrm{O}_{4}]} \]
just as with K. However, Q is evaluated with current concentrations, not those at equilibrium. When comparing Q to K, if \( Q > K \), the reaction will proceed in the reverse direction to reach equilibrium. If \( Q < K \), the reaction will move forward, and if \( Q = K \), the system is already at equilibrium. Using the reaction quotient helps in predicting the behavior of a reaction under different initial conditions and is a powerful tool in the study of dynamic chemical processes.

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Most popular questions from this chapter

Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst. For the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ equilibrium constants ( \(K_{\mathrm{p}}\) values) as a function of temperature are \(300^{\circ} \mathrm{C}, \quad 4.34 \times 10^{-3}\) \(500^{\circ} \mathrm{C}, \quad 1.45 \times 10^{-5}\) \(600^{\circ} \mathrm{C}, \quad 2.25 \times 10^{-6}\) Is the reaction exothermic or endothermic?

At a particular temperature, \(12.0\) moles of \(\mathrm{SO}_{3}\) is placed into a 3.0-L rigid container, and the \(\mathrm{SO}_{3}\) dissociates by the reaction $$ 2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) $$ At equilibrium, \(3.0\) moles of \(\mathrm{SO}_{2}\) is present. Calculate \(K\) for this reaction.

At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be \(\left[\mathrm{N}_{2}\right]=\) \(0.041 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.0078 M\), and \([\mathrm{NO}]=4.7 \times 10^{-4} M .\) Calcu- late the value of \(K\) for the reaction.

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?

Given \(K=3.50\) at \(45^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g) $$ and \(K=7.10\) at \(45^{\circ} \mathrm{C}\) for the reaction $$ 2 \mathrm{~A}(g)+\mathrm{D}(g) \rightleftharpoons \mathrm{C}(g) $$ what is the value of \(K\) at the same temperature for the reaction $$ \mathrm{C}(g)+\mathrm{D}(g) \rightleftharpoons 2 \mathrm{~B}(g) $$ What is the value of \(K_{\mathrm{p}}\) at \(45^{\circ} \mathrm{C}\) for the reaction? Starting with \(1.50\) atm partial pressures of both \(\mathrm{C}\) and \(\mathrm{D}\), what is the mole fraction of \(\mathrm{B}\) once equilibrium is reached?

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