/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 142 (A): When \(\mathrm{CaCO}_{3}(\m... [FREE SOLUTION] | 91Ó°ÊÓ

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(A): When \(\mathrm{CaCO}_{3}(\mathrm{~s})\) is heated, the loss of \(\mathrm{CO}_{2}\) (g) from the system causes the reaction to go almost to completion to leave a residue of \(\mathrm{CaO}(\mathrm{s})\) (R): Heating causes gas particles to move with more energy.

Short Answer

Expert verified
(A) is true and correctly explained by (R).

Step by step solution

01

Identify the Reaction

The decomposition reaction we're considering is \( \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \). This is a thermal decomposition reaction that occurs when calcium carbonate is heated.
02

Understand the Role of CO2

Since \( \text{CO}_2 \) is a gas, its formation and removal from the reaction mixture shift the equilibrium towards the products, effectively driving the reaction to completion according to Le Chatelier's Principle.
03

Connection Between Reaction and Heating

Heating provides the energy required to break the bonds in \( \text{CaCO}_3 \), generating \( \text{CO}_2 \) gas, which upon escaping reduces the overall pressure and drives the reaction forward.
04

Analyzing Explanation R

Statement (R) explains that heating increases the energy and movement of gas particles, enabling any formed \( \text{CO}_2 \) to be removed from the system. This removal is essential in shifting the reaction toward completion, making the role of heating pivotal.

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Key Concepts

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

Calcium Carbonate
Calcium carbonate, or \( \text{CaCO}_3 \), is a common chemical found in rocks such as limestone and chalk. It is widely used in industry, agriculture, and even as a dietary supplement. One of its notable characteristics is the ability to undergo thermal decomposition when subjected to heat. In this process, calcium carbonate breaks down into calcium oxide \( \text{CaO} \) and carbon dioxide \( \text{CO}_2 \). This decomposition reaction can be represented by the equation:
\[\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)\]When calcium carbonate is heated, the bonds holding it together have enough energy supplied to them to deteriorate, thus forming new products. This is a crucial reaction in the production of lime, an essential compound in construction and manufacturing. The process also illustrates the role of thermal energy in promoting chemical changes.
Le Chatelier's Principle
Le Chatelier's Principle helps us understand how different conditions affect the equilibrium state of a chemical reaction. When a system at equilibrium is disturbed by a change in temperature, pressure, or concentration, the system will adjust to counteract that change and restore a new equilibrium. In the case of calcium carbonate's decomposition, \( \text{CO}_2 \) is a gaseous product, and the removal of this gas from the reaction system leads to a shift in equilibrium.
Here’s how it works in this particular reaction:
  • As \( \text{CO}_2 \) gas is formed, its immediate removal from the reaction environment means less product is present in the system.
  • This removal creates an imbalance, leading to a shift in equilibrium to the right, producing more \( \text{CaO} \) and \( \text{CO}_2 \) until the reaction reaches completion.
  • Heating serves two purposes: it supplies the heat energy necessary for decomposition and facilitates the removal of \( \text{CO}_2 \) by intensifying the gas' molecular motion.
This principle effectively explains why increasing temperature in this scenario doesn’t only accelerate the reaction rate but also ensures more complete decomposition by the physical removal of \( \text{CO}_2 \).
Chemical Equilibrium
Chemical equilibrium is the state in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction, resulting in no overall change in the concentrations of reactants and products. However, not every reaction reaches an equilibrium state, especially when external conditions, such as heating in the case of calcium carbonate decomposition, are applied.
For the decomposition of calcium carbonate, the reaction is initially set to reach equilibrium under normal conditions. In a closed system without external interference, both \( \text{CaCO}_3 \) formation and decomposition occur at equal rates, maintaining equilibrium. However, when heat is applied, the following occurs:
  • The forward reaction rate increases as heat energy breaks the chemical bonds of \( \text{CaCO}_3 \).
  • The formation of \( \text{CO}_2 \) gas and its removal prevents the backward reaction from happening, effectively shifting the equilibrium towards complete decomposition.
Thus, in practical applications, chemical systems are rarely isolated, and their equilibria can be manipulated for desirable results, such as maximizing product yield in industrial processes. In the case of thermal decomposition, understanding these principles allows us to control and optimize the conditions under which reactions occur.

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Most popular questions from this chapter

Which of the following condition is not favourable for formation of more ammonia in this processI. Higher temperature II. More pressure III. More concentration of \(\mathrm{NH}_{3}\) IV. More concentration of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) a. I, II, III b. I, III C. II, III, IV d. I, III, IV

(A): The reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \leftrightarrow 2 \mathrm{NO}_{2}\) is favoured in the forward direction with increase of pressure. \((\mathbf{R})\) : The reaction is exothermic

The equilibrium constant for gaseous reaction: \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\) is 100 at \(25^{\circ} \mathrm{C}\). Consider the following statements in this regard. If the initial concentration of all the four species were \(1.0 \mathrm{M}\) each, then the equilibrium concentration of I. A would be \(0.182 \mathrm{~mol} / \mathrm{L}\) II. C would be \(0.818 \mathrm{~mol} / \mathrm{L}\) III. D would be \(1.818 \mathrm{~mol} / \mathrm{L}\) Of these statements: a. I and III are correct b. I and II are correct c. II and III are correct d. I, II and III are correct

Cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) undergoes a molecular rearrangement in the presence of \(\mathrm{AlCl}_{3}\) to form methylcyclopentane \(\left(\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right)\), according to the equation \(\mathrm{C}_{6} \mathrm{H}_{12} \rightleftharpoons \mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\) If \(\mathrm{K}_{\mathrm{C}}=0.143\) at \(25^{\circ} \mathrm{C}\) for this reaction, find the equilibrium concentrations of \(\mathrm{C}_{6} \mathrm{H}_{12}\) and \(\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\) if the initial concentrations are \(0.200 \mathrm{M}\) and \(0.100\) M respectively a. \(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right]=0.286\) and \(\left[\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right]=0.016 \mathrm{M}\) b. \(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right]=0.262\) and \(\left[\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right]=0.038 \mathrm{M}\) c. \(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right]=0.186\) and \(\left[\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right]=0.162 \mathrm{M}\) d. \(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right]=0.164\) and \(\left[\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right]=0.621 \mathrm{M}\)

The reactions in which the yield of the products can not be increased by the applications of high pressure is/are a. \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) b. \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) c. \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\) d. \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\)

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