Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The reaction \(\mathrm{Fe}^{3+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftarrows \mathrm{FeCl}^{2+}(a q)\) is endothermic. How will the equilibrium concentration of \(\mathrm{FeCl}^{2+}\) change when (a) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) is added? (b) \(\mathrm{Cl}^{-}\) is precipitated by addition of \(\mathrm{AgNO}_{3}\) ? (c) The temperature is increased? (d) A catalyst is added?

Short Answer

Expert verified
(a) Increase; (b) Decrease; (c) Increase; (d) No change.

Step by step solution

01

Understanding Reaction Type

The provided reaction \( \mathrm{Fe}^{3+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftarrows \mathrm{FeCl}^{2+}(a q) \) is endothermic, meaning it absorbs heat. According to Le Chatelier's Principle, the equilibrium will shift to counter changes like concentration or temperature.
02

Effect of Adding \( \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3} \)

Adding \( \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3} \) increases the concentration of \( \mathrm{Fe}^{3+} \) ions in the solution. The reaction will shift to the right to reduce the added \( \mathrm{Fe}^{3+} \) concentration, increasing the equilibrium concentration of \( \mathrm{FeCl}^{2+} \).
03

Effect of Precipitating \( \mathrm{Cl}^{-} \) with \( \mathrm{AgNO}_{3} \)

Adding \( \mathrm{AgNO}_{3} \) will result in the formation of \( \mathrm{AgCl} \) precipitate with \( \mathrm{Cl}^{-} \) ions, reducing the concentration of \( \mathrm{Cl}^{-} \) in solution. The equilibrium will shift to the left to increase \( \mathrm{Cl}^{-} \) ions, decreasing the concentration of \( \mathrm{FeCl}^{2+} \).
04

Effect of Increasing Temperature

Since the reaction is endothermic, increasing the temperature adds energy to the system. The equilibrium will shift to the right to absorb the extra heat, leading to an increase in the concentration of \( \mathrm{FeCl}^{2+} \).
05

Effect of Adding a Catalyst

A catalyst increases the rate at which equilibrium is reached without affecting the positions of the equilibrium. Therefore, adding a catalyst will not change the concentration of \( \mathrm{FeCl}^{2+} \) at equilibrium.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Endothermic Reactions
In chemistry, the term "endothermic" refers to reactions that absorb heat from their surroundings. This means energy is taken in during the reaction process. In the context of our reaction \(\mathrm{Fe}^{3+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftarrows \mathrm{FeCl}^{2+}(a q)\), it is described as endothermic. This attribute significantly influences how the equilibrium responds to temperature changes.

When you increase the temperature of an endothermic reaction, the system absorbs more heat. According to Le Chatelier's Principle, the equilibrium will shift in the direction that uses this added heat. For our reaction, this means a shift to the right, favoring the formation of more \(\mathrm{FeCl}^{2+}\).

To summarize, understanding whether a reaction is endothermic or exothermic is crucial when predicting the effects of temperature changes. Heating up an endothermic reaction drives the equilibrium towards the products, enhancing the concentration of products like \(\mathrm{FeCl}^{2+}\) in this scenario.
Equilibrium Shift
The concept of equilibrium shift revolves around the response of a chemical reaction to external changes, in accordance with Le Chatelier's Principle. When a system at equilibrium is subjected to a change in concentration, temperature, or pressure, it will adjust to counteract the effect of this change.

For example, in our endothermic reaction \(\mathrm{Fe}^{3+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftarrows \mathrm{FeCl}^{2+}(a q)\), several factors can cause the equilibrium to shift:
  • **Adding \(\mathrm{Fe}^{3+}\):** Results in a shift to the right to use up the excess \(\mathrm{Fe}^{3+}\), thereby increasing \(\mathrm{FeCl}^{2+}\).
  • **Removing \(\mathrm{Cl}^{-}\):** Such as by forming a precipitate with silver ions, shifts the equilibrium to the left to compensate for the lost \(\mathrm{Cl}^{-}\), reducing \(\mathrm{FeCl}^{2+}\).
  • **Increasing temperature:** Causes a shift to the right in endothermic reactions, to absorb additional heat and increase \(\mathrm{FeCl}^{2+}\) concentration.
Understanding how each change impacts the equilibrium is key to manipulating reactions effectively. Each initiative to either increase or decrease a component will be met with an equilibrium shift trying to restore balance.
Reaction Kinetics
Reaction kinetics involves the rate at which a chemical reaction proceeds. It's crucial for understanding how quickly equilibrium is reached, but it does not change the equilibrium position itself.

In our reaction, adding a catalyst could be considered for one specific purpose: speeding up the forward and reverse reactions equally, helping the system reach equilibrium faster without altering the product or reactant concentrations at equilibrium.

The phrase "adding a catalyst" should not be confused with shifting the equilibrium. Catalysts work by lowering the activation energy required for a reaction to proceed, thus increasing the reaction rate. However, they are merely speed enhancers and do not influence equilibriums' positions. Therefore, in the context of our reaction \(\mathrm{Fe}^{3+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftarrows \mathrm{FeCl}^{2+}(a q)\), the use of a catalyst will expedite reaching equilibrium but won't change the concentrations of \(\mathrm{FeCl}^{2+}\) at that equilibrium state.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is meant by the term heat of reaction? What other name is a synonym for this term?

Does entropy increase or decrease in the following processes? (a) Polymeric complex carbohydrates are metabolized by the body, converted into smaller simple sugars. (b) Steam condenses on a glass surface. (c) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\)

For each of the following processes, specify whether entropy increases or decreases. Explain each of your answers. (a) Assembling a jigsaw puzzle $$ \text { (b) } \mathrm{I}_{2}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{IF}_{3}(g) $$ (c) A precipitate forming when two solutions are mixed (d) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) 6 \longrightarrow \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (e) \(\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\) $$ \text { (f) } \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{NaCl}(a q) \longrightarrow $$ \(\mathrm{PbCl}_{2}(s)+2 \mathrm{NaNO}_{3}(a q)\)

What effect do the listed changes have on the position of the equilibrium in the reaction of carbon with hydrogen? $$ \mathrm{C}(s)+2 \mathrm{H}_{2}(g) \rightleftarrows \mathrm{CH}_{4}(g) \quad \Delta H=-75 \mathrm{~kJ} / \mathrm{mol} $$ (a) Increasing temperature (b) Increasing pressure by decreasing volume (c) Allowing \(\mathrm{CH}_{4}\) to escape continuously from the reaction vessel

Which of the following processes results in an increase in entropy of the system? (a) A drop of ink spreading out when it is placed in water (b) Steam condensing into drops on windows (c) Constructing a building from loose bricks

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free