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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 N鈧侽鈧 and NO鈧 are approximately 0.0988 M and 2.4 脳 10鈦宦 M, respectively.

Step by step solution

01

Calculate the initial concentration of N鈧侽鈧 (g)

To calculate the initial concentration of N鈧侽鈧 (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 N鈧侽鈧 and NO鈧: $$ [\mathrm{N}_{2} \mathrm{O}_{4}]_{equilibrium} = 0.1-1.2 \times 10^{-3} \approx 0.0988 \mathrm{\,M} $$ $$ [\mathrm{NO}_{2}]_{equilibrium} = 2x = 2 \times (1.2 \times 10^{-3}) \approx 2.4 \times 10^{-3} \mathrm{\,M} $$ At equilibrium, the concentrations of N鈧侽鈧 and NO鈧 are approximately 0.0988 M and 2.4 脳 10鈦宦 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 N鈧侽鈧 in a 10.0 L vessel. Initially, we have no NO鈧, so its concentration is 0. We represent the initial concentration of N鈧侽鈧 as 0.1 M and NO鈧 as 0 M. As the reaction proceeds towards equilibrium, N鈧侽鈧 decomposes into NO鈧, 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 N鈧侽鈧 will be '0.1 - x' M and for NO鈧, 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 N鈧侽鈧 and NO鈧 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鈦烩伔, indicating the equilibrium heavily favors the reactants, N鈧侽鈧.
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

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?

At a particular temperature, \(8.1\) moles of \(\mathrm{NO}_{2}\) gas is placed in a 3.0-L container. Over time the \(\mathrm{NO}_{2}\) decomposes to NO and \(\mathrm{O}_{2}\) : $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ At equilibrium the concentration of \(\mathrm{NO}(g)\) was found to be \(1.4 \mathrm{~mol} / \mathrm{L}\). Calculate the value of \(K\) for this reaction.

The following equilibrium pressures were observed at a certain temperature for the reaction $$ \begin{array}{c} \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \\\ P_{\mathrm{NH}_{3}}=3.1 \times 10^{-2} \mathrm{~atm} \\ P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{~atm} \\ P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{~atm} \end{array} $$ Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If \(P_{\mathrm{N}_{2}}=0.525 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=0.0167 \mathrm{~atm}\), and \(\underline{P_{\mathrm{H}}}=0.00761\) atm, does this represent a system at equilibrium?

Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) . \mathrm{A}\) friend asks the following: "I know we have been told that if a mixture of \(A, B, C\), and \(D\) is at equilibrium and more of \(A\) is added, more \(\mathrm{C}\) and \(\mathrm{D}\) will form. But how can more \(\mathrm{C}\) and \(\mathrm{D}\) form if we do not add more \(\mathrm{B}\) ?' What do you tell your friend?

The value of the equilibrium constant \(K\) depends on which of the following (more than one answer may be correct)? a. the initial concentrations of the reactants b. the initial concentrations of the products c. the temperature of the system d. the nature of the reactants and products Explain.

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