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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.

Short Answer

Expert verified
The value of the equilibrium constant (K) depends on the temperature of the system (option c) and the nature of the reactants and products involved in the reaction (option d). Initial concentrations of reactants and products (options a and b) only affect the position of equilibrium but not the value of K.

Step by step solution

01

Understanding the meaning of the equilibrium constant

The equilibrium constant (K) is a numerical value that expresses the ratio of product concentrations to reactant concentrations when a chemical reaction reaches equilibrium. It quantifies the tendency of a reaction to proceed towards completion at a given temperature.
02

Addressing initial concentrations of reactants and products

The initial concentrations of reactants and products affect the position of equilibrium, not the equilibrium constant. Changes in initial concentrations can cause the reaction to shift towards either the reactants' side or the products' side until a new equilibrium position is established. However, the value of the equilibrium constant (K) for a particular reaction remains constant at a given temperature, regardless of the initial concentrations of reactants and products. Therefore, options a and b are incorrect.
03

Analyzing the effect of temperature on the equilibrium constant

The value of the equilibrium constant (K) depends on the temperature of the system. An increase in temperature can cause the value of K to increase or decrease, depending on whether the reaction is endothermic (where heat is absorbed) or exothermic (where heat is released). For endothermic reactions, an increase in temperature will cause K to increase, whereas for exothermic reactions, an increase in temperature will cause K to decrease. Thus, option c is correct.
04

Assessing the impact of the nature of reactants and products on the equilibrium constant

The equilibrium constant (K) depends on the nature of the reactants and products involved in the chemical reaction. Different reactions involving different substances will have different equilibrium constants because the K value is related to the intrinsic properties of the substances and how they interact. Therefore, the value of K is specific to a given reaction and cannot be applied to another reaction with different reactants and products. Hence, option d is correct.
05

Conclusion

In conclusion, the value of the equilibrium constant (K) depends on the temperature of the system (option c) and the nature of the reactants and products involved in the reaction (option d). It does not depend on the initial concentrations of reactants and products (options a and b).

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, often represented as \(K\), is a fundamental concept in chemistry. It quantifies the ratio of concentrations of products to reactants at equilibrium for a given chemical reaction. This ratio reflects the extent to which a reaction will proceed before reaching a stable state. Once equilibrium is achieved, the forward and reverse reactions occur at the same rate, maintaining constant concentrations.

The expression for the equilibrium constant depends on the balanced chemical equation of the reaction. For a general reaction:
  • \(aA + bB \rightleftharpoons cC + dD\)
\(K\) is calculated as:
  • \(K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\)
Here, the square brackets indicate concentration molarity, and the lowercase letters represent stoichiometric coefficients of the reactants (\(A\) and \(B\)) and products (\(C\) and \(D\)). It is important to note that \(K\) is only defined at equilibrium and only changes with temperature, as seen in the connections between temperature and reaction dynamics.
Temperature Effects on Equilibrium
Temperature plays a pivotal role in determining the value of the equilibrium constant \(K\). While \(K\) remains unaffected by concentrations of reactants or products initially, it is sensitive to temperature changes. The outcome of these changes depends on whether a chemical reaction is endothermic or exothermic.
  • **Endothermic Reactions:** For reactions that absorb heat, an increase in temperature results in a higher equilibrium constant \(K\). This indicates a greater tendency for the reaction to produce more products, thus shifting equilibrium to the right.
  • **Exothermic Reactions:** Conversely, in reactions that release heat, an increase in temperature reduces the value of \(K\). Here, it suggests a tendency to form more reactants, shifting equilibrium to the left.
This dependence on temperature is attributed to the balance of energy changes within the reaction system. Understanding this relationship is crucial for predicting how equilibrium will adjust to changing environmental conditions.
Nature of Reactants and Products
The intrinsic properties of the reactants and products are also key factors that determine the equilibrium constant \(K\). Different chemical species have unique energies and interact through varying mechanisms. Therefore, even if two reactions have similar conditions, their equilibrium constants can differ significantly due to the nature of substances involved.

The chemical identity of reactants and products influences how they collide, react, and achieve equilibrium. Factors such as:
  • Molecular structure
  • Phase (solid, liquid, gas)
  • Bond strengths and enthalpies
impact the overall energy balance and reaction pathway. This inherent characteristic of substances makes \(K\) specific to a particular reaction. It means that \(K\) values cannot be transferred between reactions with different chemical compositions, providing a specific 'reaction fingerprint' for each unique chemical equation.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}}=5.3 \times 10^{5}\) for the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ When a certain partial pressure of \(\mathrm{NH}_{3}(g)\) is put into an otherwise empty rigid vessel at \(25^{\circ} \mathrm{C},\) equilibrium is reached when \(50.0 \%\) of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?

Le Châtelier's principle is stated (Section \(12-7\) ) as follows: "If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change." The system \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in \(\mathrm{H}_{2}\) concentration and an increase in \(\mathrm{NH}_{3}\) concentration. In the experiment the volume is assumed to be constant. On the other hand, if \(\mathrm{N}_{2}\) is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of \(\mathrm{NH}_{3}\) actually could decrease, and the concentration of \(\mathrm{H}_{2}\) would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.

Novelty devices for predicting rain contain cobalt(II) chloride and are based on the following equilibrium:$$\mathrm{CoCl}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s)$$ Purple Pink.What color will such an indicator be if rain is imminent?

Calculate a value for the equilibrium constant for the reaction $$\mathbf{O}_{2}(g)+\mathbf{O}(g) \rightleftharpoons \mathbf{O}_{3}(g)$$.given $$\begin{aligned}& \mathrm{NO}_{2}(g) \stackrel{h v}{\rightleftharpoons} \mathrm{NO}(g)+\mathrm{O}(g) & & K=6.8 \times 10^{-49} \\\\\mathrm{O}_{3}(g)+\mathrm{NO}(g) & \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) & & K=5.8 \times 10^{-34}\end{aligned}$$.(Hint: When reactions are added together, the equilibrium expressions are multiplied.) (Hint: When reactions are added together, the equilibrium expressions are multiplied.)

Consider the decomposition equilibrium for dinitrogen pentoxide: $$2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2(g)+\mathrm{O}_{2}(g)$$.At a certain temperature and a total pressure of 1.00 atm, the \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(0.50 \%\) decomposed (by moles) at equilibrium. a. If the volume is increased by a factor of \(10.0,\) will the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposed at equilibrium be greater than, less than, or equal to 0.50\%? Explain your answer. b. Calculate the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) that will be decomposed at equilibrium if the volume is increased by a factor of 10.0

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