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Write the expression for \(K_{\mathrm{eq}}\) for the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g)\).

Short Answer

Expert verified
The equilibrium constant expression, \(K_{eq}\), for the reaction \(\mathrm{N}_{2}\mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g)\) is: \[K_{eq} = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]}\]

Step by step solution

01

(Step 1: Identify the Reaction)

(The given chemical reaction is: \(\mathrm{N}_{2}\mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g)\) This is a reversible reaction where \(\mathrm{N}_{2}\mathrm{O}_{4}(g)\) is in equilibrium with \(2 \mathrm{NO}_{2}(g)\).)
02

(Step 2: Define Reaction Quotient (Q))

(We first define the reaction quotient, Q, which is similar to the equilibrium constant expression but involves the concentrations (or partial pressures) of the reactants and products at any point in time. For the given reaction, the Q expression is: \[Q = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]}\] Here, the square brackets denote the molar concentration (in moles per liter) of each substance. Note that we use a mole ratio to express the concentration of products over reactants, which is \(2^\text{nd}\) for the product \(2\) NO2(g), and the power is denoted by the coefficient of that reactant or product.)
03

(Step 3: Define Equilibrium Constant (K_eq))

(When the reaction reaches equilibrium, the concentrations of all the reactants and products won't change anymore, but they won't necessarily be equal. At this point, the reaction quotient (Q) will become the equilibrium constant (K_eq). Because the process remains the same, the expression for K_eq will also be the same as Q: \[K_{eq} = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]}\] Now we have the final expression for the equilibrium constant for the given chemical reaction.)

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Key Concepts

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

Chemical Equilibrium
The fascinating dance of chemical reactions often leads to a state called \textbf{chemical equilibrium}. Imagine a crowded dance floor where the movement of dancers striking a balance – no net change in the number of dancers, even though they're constantly in motion. Similarly, in a chemical reaction at equilibrium, such as \(\mathrm{N}_{2}\mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g)\), there's a dynamic balance between reactants converting to products and products reverting to reactants. The key to equilibrium is that, despite ongoing change at the molecular level, the macroscopic properties, such as concentration, color, and pressure, remain constant over time.

At this magical point, the rate of the forward reaction equals the rate of the reverse reaction. The ratio of the product concentrations to the reactant concentrations remains stable, leading to the equilibrium constant expression (K_eq). This constant offers a way of determining whether a system is at equilibrium and, if so, in which direction it is most likely to proceed to reach that state.
Reaction Quotient (Q)
As a 'snapshot' of a reaction's progress, the \textbf{reaction quotient (Q)} is like a chemical 'status update' that tells us how far or close a reaction is from reaching equilibrium. It’s calculated using the same formula as \(K_{eq}\) but with one crucial difference – while \(K_{eq}\) is the constant when a reaction is at equilibrium, Q refers to concentrations at any moment before the system has reached that state.

Think of Q as a GPS for chemists. For our reaction \(\mathrm{N}_{2}\mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g)\), the expression for Q is \[Q = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]}\], where the numbers inside the square brackets represent the molar concentrations. If Q equals \(K_{eq}\), the system is at equilibrium. If Q is different, it's not at equilibrium, and the reaction will shift to favor either the forward or reverse process to achieve balance, depending on whether Q is greater or less than \(K_{eq}\).
Molar Concentration
When we measure the 'strength' or 'intensity' of a chemical species in a solution, we rely on a term called \textbf{molar concentration}, often represented by square brackets around a substance. For example, \[\mathrm{NO}_2\] signifies the concentration of \(\mathrm{NO}_2\) measured in moles per liter (mol/L). Just like counting the number of coffee beans in a bag gives you an idea of how strong your brew might be, measuring the molar concentration tells you about the 'potency' of a reactant or product in a chemical reaction.

However, it's not just the amount that matters – it's also the proportion. When we construct the equilibrium constant expression (as in step 2 for \(K_{eq}\)), we must consider the stoichiometry of the reaction. This means for our reversible dance between \(\mathrm{N}_{2}\mathrm{O}_{4}\) and \(\mathrm{NO}_2\), the concentration of \(\mathrm{NO}_2\) is squared because the chemical equation shows us that two moles of \(\mathrm{NO}_2\) are involved for every one mole of \(\mathrm{N}_{2}\mathrm{O}_{4}\). By knowing these concentrations, it becomes possible to understand and predict the behavior and progress of the reaction towards equilibrium.

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Most popular questions from this chapter

Suppose we have an equilibrium mixture of reactants and products for the reaction \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows \mathrm{PCl}_{5}(g)\) Predict the direction in which the reaction will shift when: (a) Chlorine \(\left(\mathrm{Cl}_{2}\right)\) gas is added. (b) Chlorine gas is removed. (c) \(\mathrm{PCl}_{5}\) is added. (d) \(\mathrm{PCl}_{3}\) is removed. (e) \(\mathrm{H}_{2}\) gas is added. (Assume the \(\mathrm{H}_{2}\) does not react with any reactant or product.)

A certain reaction has a \(K_{\text {eq }}\) value of \(1.5 \times 10^{-6}\). (a) Would this be a practical reaction from which to isolate pure product? Explain your answer.

Write the equilibrium constant expression for the reaction \(\mathrm{CaCO}_{3}(s) \leftrightarrows \mathrm{CaO}(s)+\) \(\mathrm{CO}_{2}(g)\)

Some liquid water is placed in a room-temperature container and sealed. After some time, the level of the liquid drops by a small amount. After that, the water level remains constant.

\(K_{\text {eq }}=3.9 \times 10^{-11}\) for the dissolution of calcium fluoride in water: \(\mathrm{CaF}_{2}(s) \rightleftarrows \mathrm{Ca}^{2+}(a q)+2 \mathrm{~F}^{-}(a q)\) (a) What is another name for \(K_{\mathrm{eq}}\) for this reaction? (b) If the equilibrium calcium ion concentration in a saturated aqueous solution of calcium fluoride is \(3.3 \times 10^{-4} \mathrm{M}\), what is the equilibrium fluoride ion concentration? (c) Which is larger, the rate constant for the forward reaction or the rate constant for the reverse reaction? (d) Which is larger, \(E_{a}\) for the forward reaction or \(E_{\mathrm{a}}\) for the reverse reaction? (e) Which is larger, the rate of the forward reaction or the rate of the reverse reaction? (f) For lithium carbonate, \(K_{\mathrm{sp}}=0.0011\). Write the balanced chemical equation and the equilibrium expression for the dissolution of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) in water. (g) Which is more soluble in water, calcium fluoride or lithium carbonate?

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