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Write the expression for \(K_{\text {eq }}\) for the reaction \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftarrows 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)

Short Answer

Expert verified
The expression for the equilibrium constant, \(K_{eq}\), for the given reaction is: \[K_{eq} = \frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}\]

Step by step solution

01

Identify the reactants and products

In the given reaction, the reactants are: - Ammonia (NH3) with a coefficient of 4 - Oxygen (O2) with a coefficient of 5 The products are: - Nitric oxide (NO) with a coefficient of 4 - Water vapor (H2O) with a coefficient of 6
02

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

Using the general formula for the equilibrium constant, we have \[K_{eq} = \frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}\] This gives us the expression for the equilibrium constant for the given reaction.

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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 state in a chemical reaction where the rates of the forward and reverse reactions are equal. This balance does not mean the reactants and products are in equal amounts, but rather that their concentrations remain constant over time.
At this point, no observable changes occur in the concentrations of reactants and products. Chemical equilibrium is dynamic because the reactions continue to occur, but with no net change in the system.
The equilibrium constant, denoted as \( K_{eq} \), is a value that expresses the relationship between the concentrations of products and reactants at equilibrium. The size of \( K_{eq} \) gives insight into the position of equilibrium:
  • If \( K_{eq} \) is much greater than 1, the equilibrium lies to the right, favoring product formation.
  • If \( K_{eq} \) is much less than 1, the equilibrium lies to the left, favoring reactants.
Reaction Coefficients
In a balanced chemical equation, the numbers in front of the chemical formulas are called the reaction coefficients. These coefficients indicate the molar ratio in which reactants and products participate in the reaction.
For the equilibrium expression, these coefficients convert into exponents.
  • In the example reaction \(4 \text{NH}_3 (g) + 5 \text{O}_2 (g) \rightleftarrows 4 \text{NO} (g) + 6 \text{H}_2 \text{O} (g)\), the coefficients are 4, 5, 4, and 6, respectively.
  • In the equilibrium constant expression \(K_{eq} = \frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}\), each concentration term is raised to the power of its reaction coefficient.
This reflects the intuitive relationship that larger coefficients imply a greater involvement in the reaction, affecting equilibrium calculations proportionally.
Le Chatelier's Principle
Le Chatelier's principle is a fundamental concept in chemistry that describes how a system at equilibrium responds to external changes. When a change, such as concentration, pressure, or temperature, is applied to a system in equilibrium, the system will adjust to counteract the change and re-establish equilibrium.
This principle provides a way to predict the direction of the shift in equilibrium:
  • If additional reactants are added, the system will shift to produce more products.
  • If products are removed or reactants are decreased, the system will shift towards producing more products.
  • Changes in temperature can either favor the exothermic or endothermic direction depending on whether the temperature is increased or decreased.
Understanding Le Chatelier's principle helps in controlling the conditions for industrial chemical reactions to optimize yield and efficiency.

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Most popular questions from this chapter

What effect does a catalyst have on: (a) The position of equilibrium for a reaction? (b) The value of the equilibrium constant? (c) The ratio of \(k_{\mathrm{f}} / k_{\mathrm{r}} ?\) 14.86 What does a catalyst do to the time it takes for a reaction to reach equilibrium? Explain how it does this.

At \(25^{\circ} \mathrm{C}\), the solubility in water of the moderately soluble salt silver acetate, \(\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) is \(10.6 \mathrm{~g} / \mathrm{L}\). (a) Write the chemical equation for the dissolving of silver acetate in water. (b) Write the \(K_{\mathrm{sp}}\) expression for silver acetate. (c) Calculate the value of \(K_{\mathrm{sp}}\) (show your work).

\(K_{\text {eq }}=3.9 \times 10^{-11}\) 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?

Consider the two curves, one showing supply of tapes, the other showing demand for tapes. At what price per quantity is equilibrium established?

Write the balanced chemical equation for the reaction that goes with the equilibrium constant \(K_{\mathrm{eq}}=\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]^{2} \times\left[\mathrm{Cl}_{2}\right]^{2}}{[\mathrm{HCl}]^{4} \times\left[\mathrm{O}_{2}\right]}\)

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