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The equilibrium concentrations for the reaction \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\) at \(2000^{\circ} \mathrm{C}\) are \(\left[\mathrm{N}_{2}\right]=\) \(0.25 \mathrm{M} ;\left[\mathrm{O}_{2}\right]=1.2 \mathrm{M} ;[\mathrm{NO}]=0.011 \mathrm{M} .\) What is the value of \(K_{\mathrm{eq}}\) for this reaction?

Short Answer

Expert verified
The value of the equilibrium constant, \(K_{\text{eq}}\), for the reaction \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\) at \(2000^{\circ}\mathrm{C}\) is \(4.03 \times 10^{-4}\).

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation is given as: \[\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\]
02

Write the expression for the equilibrium constant \(K_{\text{eq}}\)

The equilibrium constant for the given reaction is the ratio of the product of the molar concentrations of the products to the product of the molar concentrations of the reactants, each raised to the power of the stoichiometric coefficients in the balanced equation: \[K_{\text{eq}} = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_{2}] \cdot [\mathrm{O}_{2}]}\]
03

Insert the equilibrium concentrations into the formula

We are given the following equilibrium concentrations: \([\mathrm{N}_{2}]=0.25\,\mathrm{M}\), \([\mathrm{O}_{2}]=1.2\,\mathrm{M}\), and \([\mathrm{NO}]=0.011\,\mathrm{M}\). Substitute these values into the equilibrium constant expression: \[K_{\text{eq}} =\frac{(0.011)^2}{(0.25) \cdot (1.2)}\]
04

Calculate the value of \(K_{\text{eq}}\)

Perform the calculation to determine the value of the equilibrium constant: \[K_{\text{eq}} =\frac{(0.011)^2}{(0.25) \cdot (1.2)} =\frac{0.000121}{0.3} = 4.03 \times 10^{-4}\] The value of the equilibrium constant, \(K_{\text{eq}}\), for this reaction at \(2000^{\circ}\mathrm{C}\) is \(4.03 \times 10^{-4}\).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a fascinating concept in chemistry that describes a situation where a chemical reaction and its reverse occur at the same rate. When this happens, the concentrations of reactants and products remain constant over time, even though both reactions are still active. This state does not mean the reactants and products are equal, but rather their rates of formation and decomposition are balanced.
  • It is dynamic, which means the reactions are ongoing.
  • Equilibrium can shift with changes in concentration, pressure, or temperature.
  • The notion of equilibrium is crucial for understanding how reactions behave under different conditions.
One key parameter related to chemical equilibrium is Keq, the equilibrium constant. This constant allows chemists to predict how far a reaction will proceed to either the left (reactants) or right (products). Understanding this helps in managing industrial chemical processes effectively.
Reaction Concentrations
Reaction concentrations play a pivotal role in defining the position of equilibrium in a chemical reaction. At equilibrium, the rates of both the forward and reverse reactions are equal, leading to constant concentrations of reactants and products.
It is important to measure these concentrations accurately as they are used to determine the equilibrium constant (Keq). In the original exercise, we learned how those values are critical in deriving Keq using the expression:
\[K_{eq} = \frac{[\text{NO}]^2}{[\text{N}_2] \cdot [\text{O}_2]}\]The reaction concentrations must reflect the stoichiometric coefficients given in the balanced equation. Misjudging these can lead to errors in equilibrium constant calculations and hinder our understanding of a system's behavior. Utilization of these concentrations aids chemists in predicting reaction sway and can influence decisions like altering reactant feeds or reaction conditions.
Stoichiometry
Stoichiometry is the area of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is vital for understanding how much of each substance is consumed or produced during a reaction.
In the context of equilibrium, the stoichiometric coefficients from the balanced chemical equation become exponents in the Keq expression. For example, in the equation
\[\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\]The coefficient '2' in front of NO is used as an exponent in the Keq formula. Thus, stoichiometry not only guides how equations are balanced but also directly influences the calculation and interpretation of equilibrium constants.
  • Correct stoichiometric coefficients ensure accurate equilibrium calculations.
  • They help determine reaction yields.
  • They can hint at potential reaction pathways.
Understanding stoichiometry is essential for manipulating chemical reactions to achieve desired outcomes.
Temperature Influence on Equilibrium
Temperature is a crucial factor that can significantly influence chemical equilibrium. According to Le Chatelier's principle, if an equilibrium system is subjected to a change in temperature, the system will adjust to counteract that change and restore balance.
Increasing temperature generally favors the endothermic direction of a reaction, whereas decreasing temperature favors the exothermic direction. In the original exercise, knowing the equilibrium was at a specified temperature (2000°C) was vital for context.
  • Temperature changes can shift equilibrium positions, altering concentrations of reactants and products.
  • Understanding this concept helps in controlling industrial processes, like the synthesis of ammonia.
  • Temperature effects must be considered when designing reactions to avoid unwanted shifts that could affect yield.
Hence, temperature doesn't just change the rate of reactions but can impact the actual equilibrium state and the value of Keq. It’s an essential consideration in managing reaction conditions to optimize desired product formation.

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Most popular questions from this chapter

The saturation solubility of \(\mathrm{Ag}_{2} \mathrm{~S}\) at \(25^{\circ} \mathrm{C}\) is \(1.14 \times 10^{-17} \mathrm{M}\). What are the equilibrium concentrations of the cation and anion?

Consider the reaction \(\mathrm{SnO}_{2}(s)+2 \mathrm{H}_{2}(g) \rightleftarrows \mathrm{Sn}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) run in an explosion-proof sealed vessel. (a) Running the reaction in a sealed vessel allows equilibrium to be established. Explain why. (b) Express the concentration of \(\mathrm{H}_{2}(g)\) in terms of \(K_{\text {eq }}\) (c) The equilibrium constant for this reaction decreases as the reaction mixture is heated. Which way does the equilibrium shift? (d) Is the reaction exothermic or endothermic? Explain. (e) A new employee at a chemical company adds additional \(\mathrm{SnO}_{2}(s)\) to the reaction in order to drive its position of equilibrium further to the right. Will this get her a promotion or get her fired? Explain.

Hydrochloric acid is added to an aqueous solution of silver nitrate. (a) What precipitate forms? (b) The equilibrium concentration of \(\mathrm{Cl}^{-}(a q)\) is \(2.0 \times 10^{-3} \mathrm{M}\). What is the equilibrium concentration of \(\mathrm{Ag}^{+}(a q) ?\)

If \(k_{\mathrm{f}}>k_{\mathrm{r}}\), will \(K_{\mathrm{eq}}\) be less than 1 or greater than 1? Explain your answer.

The water in a beaker of water left in a room will slowly evaporate until the beaker is dry. However, place that same beaker in a sealed box and the water level in the beaker will drop a bit but then remain constant. Is the latter case an example of equilibrium? Explain your answer.

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