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List four factors that can shift the position of an equilibrium. Only one of these factors can alter the value of the equilibrium constant. Which one is it?

Short Answer

Expert verified
The four factors that can shift the position of an equilibrium are changes in concentration, pressure, temperature, and the addition of a catalyst. However, only changes in temperature can alter the value of the equilibrium constant.

Step by step solution

01

Identifying the four factors

The four primary factors that can affect the position of an equilibrium include: Changes in concentration, Changes in temperature, Changes in pressure, Addition of a catalyst.
02

Analyzing each factor: Concentration

Changes in the concentration of either reactants or products can shift the position of the equilibrium according to Le Chatelier's principle. If the concentration of a reactant is increased, the system will shift to the right, favoring the forward reaction and thus producing more products. Similarly, if the concentration of a product is increased, the system will shift to the left, favoring the reverse reaction and producing more reactants. However, changes in concentration do not influence the equilibrium constant.
03

Analyzing each factor: Pressure

Changes in pressure can also shift the position of the equilibrium, especially for reactions involving gases. Increasing the pressure results in the system shifting in the direction that reduces the pressure, i.e., towards the side with fewer gas molecules (lower volume). Reducing the pressure shifts the equilibrium towards the side with more gas molecules (higher volume). Still, changes in pressure do not affect the equilibrium constant.
04

Analyzing each factor: Temperature

Changes in temperature can shift the equilibrium and also change the value of the equilibrium constant, making this the answer to the second part of the question. An increase in temperature for an endothermic reaction (one that absorbs heat) will shift the equilibrium towards the products and increase K. On the other hand, for an exothermic reaction (a reaction that releases heat), an increase in temperature will shift the equilibrium towards the reactants and decrease K.
05

Analyzing each factor: Catalyst

The addition of a catalyst only speeds up the rate at which equilibrium is reached. It does not shift the position of the equilibrium nor does it change the value of the equilibrium constant.

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry. It helps predict how a change in conditions can affect the position of a chemical equilibrium. Simply put, if a system at equilibrium experiences a change in concentration, pressure, or temperature, it will adjust to counteract the change and restore a new equilibrium. This principle provides critical insight into the dynamic nature of chemical reactions.

Here's how the principle works in simple terms:
  • If the concentration of reactants is increased, the equilibrium shifts towards the products to consume the extra reactants.
  • If the concentration of products is increased, the equilibrium shifts towards the reactants to reduce the extra products.
  • In a gaseous equilibrium, if pressure is increased, the system shifts towards the side with fewer gas molecules to reduce pressure.
  • If temperature is increased, the system favors the endothermic direction to absorb the additional heat.
Le Chatelier's Principle is a powerful tool for predicting the qualitative outcome of perturbations in a chemical system.
Equilibrium Constant
The equilibrium constant, denoted as \( K \), is a crucial concept that characterizes the balance point of a reversible chemical reaction at a particular temperature. It is expressed as a ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients.

Mathematically, it is represented as:\[K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]where \([A]\), \([B]\), \([C]\), and \([D]\) are the molar concentrations of the reactants and products, and \(a\), \(b\), \(c\), and \(d\) are their respective coefficients in the balanced chemical equation.

Temperature is the only factor that can alter the value of \( K \). For endothermic reactions, an increase in temperature generally increases \( K \), meaning more products are favored. Conversely, for exothermic reactions, increasing temperature decreases \( K \), favoring reactants.
Factors Affecting Equilibrium
Several factors can influence the state of equilibrium in a chemical reaction, affecting either the position of the equilibrium or the equilibrium constant itself. The key factors are:

  • **Concentration**: Modifying the concentration of reactants or products shifts the equilibrium according to Le Chatelier's Principle. However, it does not change the equilibrium constant \( K \).
  • **Pressure**: This affects equilibria involving gaseous reactants or products. Increasing the pressure will shift the equilibrium toward the side with fewer gas molecules, without changing \( K \).
  • **Temperature**: Temperature changes can both shift equilibrium positions and alter \( K \). This is unique among the factors; higher temperatures favor endothermic processes and lower \( K \) for exothermic ones.
  • **Catalysts**: While catalysts accelerate the rate at which equilibrium is reached, they do not affect the position of equilibrium or change the value of \( K \).
Understanding these factors is essential for controlling chemical reactions in industrial processes, laboratories, and nature.

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Most popular questions from this chapter

When dissolved in water, glucose (corn sugar) and fructose (fruit sugar) exist in equilibrium as follows: fructose \(\rightleftharpoons\) glucose A chemist prepared a \(0.244 M\) fructose solution at \(25^{\circ} \mathrm{C}\). At equilibrium, it was found that its concentration had decreased to \(0.113 M .\) (a) Calculate the equilibrium constant for the reaction. (b) At equilibrium, what percentage of fructose was converted to glucose?

Consider this equilibrium reaction in a closed container: $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ What will happen if the following occurs? (a) The volume is increased. (b) Some \(\mathrm{CaO}\) is added to the mixture. (c) Some \(\mathrm{CaCO}_{3}\) is removed. (d) Some \(\mathrm{CO}_{2}\) is added to the mixture. (e) A few drops of a \(\mathrm{NaOH}\) solution are added to the mixture. (f) A few drops of a \(\mathrm{HCl}\) solution are added to the mixture (ignore the reaction between \(\mathrm{CO}_{2}\) and water). (g) Temperature is increased.

About 75 percent of hydrogen for industrial use is produced by the steam- reforming process. This process is carried out in two stages called primary and secondary reforming. In the primary stage, a mixture of steam and methane at about 30 atm is heated over a nickel catalyst at \(800^{\circ} \mathrm{C}\) to give hydrogen and carbon monoxide: $$\begin{array}{r}\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \\\\\Delta H^{\circ}=260 \mathrm{~kJ} /\mathrm{mol}\end{array}$$ The secondary stage is carried out at about \(1000^{\circ} \mathrm{C}\), in the presence of air, to convert the remaining methane to hydrogen: $$\begin{array}{r}\mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \\\\\Delta H^{\circ}=35.7 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ (a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stage? (b) The equilibrium constant \(K_{\mathrm{c}}\) for the primary stage is 18 at \(800^{\circ} \mathrm{C}\). (i) Calculate \(K_{P}\) for the reaction. (ii) If the partial pressures of methane and steam were both 15 atm at the start, what are the pressures of all the gases at equilibrium?

At room temperature, solid iodine is in equilibrium with its vapor through sublimation and deposition (see Section 11.8). Describe how you would use radioactive iodine, in either solid or vapor form, to show that there is a dynamic equilibrium between these two phases.

Photosynthesis can be represented by $$\begin{array}{r}6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \\\\\Delta H^{\circ}=2801 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ Explain how the equilibrium would be affected by the following changes: (a) Partial pressure of \(\mathrm{CO}_{2}\) is increased. (b) \(\mathrm{O}_{2}\) is removed from the mixture. (c) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (glucose) is removed from the mixture. (d) More water is added. (e) A catalyst is added. (f) Temperature is decreased.

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