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When 2 mol of potassium chlorate crystals decompose to potassium chloride crystals and oxygen gas at constant temperature and pressure, \(78.0 \mathrm{~kJ}\) of heat is given off. Write a thermochemical equation for this reaction.

Short Answer

Expert verified
The thermochemical equation is: \( 2 \text{KClO}_3 (s) \rightarrow 2 \text{KCl} (s) + 3 \text{O}_2 (g) \), \( \Delta H = -78.0 \text{ kJ} \).

Step by step solution

01

Understanding the Chemical Equation

The chemical reaction for the decomposition of potassium chlorate (\( \text{KClO}_3 \)) into potassium chloride (\( \text{KCl} \)) and oxygen gas (\( \text{O}_2 \)) is: \[ 2 \text{KClO}_3 (s) \rightarrow 2 \text{KCl} (s) + 3 \text{O}_2 (g) \]. This equation shows that 2 moles of potassium chlorate decompose to produce 2 moles of potassium chloride and 3 moles of oxygen gas.
02

Identify the Heat Involved

According to the problem statement, 78.0 kJ of heat is given off during this decomposition. This information tells us that the reaction is exothermic, as heat is released.
03

Writing the Thermochemical Equation

A thermochemical equation includes the balanced chemical equation and the enthalpy change (\( \Delta H \)) of the reaction. Since heat is given off and the reaction is exothermic, \( \Delta H \) is negative. We write this as: \[ 2 \text{KClO}_3 (s) \rightarrow 2 \text{KCl} (s) + 3 \text{O}_2 (g) \quad \Delta H = -78.0 \text{ kJ} \]

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Key Concepts

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

Chemical Decomposition
Chemical decomposition, also known as chemical breakdown, is a process where a compound breaks apart into two or more smaller compounds or elements. It is akin to taking a large molecule and breaking it down into its building blocks.

In the context of our exercise, the compound involved is potassium chlorate (\(\text{KClO}_3\)). When it decomposes, it breaks down into potassium chloride (\(\text{KCl}\)) and oxygen gas (\(\text{O}_2\)). Thus, during the decomposition:
  • 2 moles of potassium chlorate yields 2 moles of potassium chloride and 3 moles of oxygen gas.
Chemical decomposition reactions are essential in various fields, including chemistry and industrial manufacturing, as they allow the extraction of useful materials and substances from complex compounds.
Exothermic Reactions
Exothermic reactions are a type of chemical reaction where energy is released to the surroundings, usually in the form of heat. This results in the surrounding temperatures increasing. You'll often feel these reactions as they release heat.

The decomposition of potassium chlorate in our exercise is exothermic because it releases 78.0 kJ of energy. This release of heat is typically indicated in thermochemical equations with a negative enthalpy change (\(\Delta H\)).
  • Exothermic reactions are common in combustion processes, such as when burning wood.
  • They are also found in chemical hand warmers or self-heating cans.
Recognizing exothermic reactions is crucial as they often provide energy for further reactions or applications.
Enthalpy Change
Enthalpy change (\(\Delta H\)) is the heat change that occurs during a chemical reaction at constant pressure. It is a measure of the total energy gained or lost by a system. In essence, it tells us how much heat is absorbed or released during a reaction.

In exothermic reactions, like the decomposition of potassium chlorate, the enthalpy change is negative because heat is released. This concept is crucial in understanding how energy flows in chemical processes, which can affect everything from the stability of products to the feasibility of industrial processes.
  • Negative \(\Delta H\) values indicate energy release (exothermic).
  • Positive \(\Delta H\) values show energy absorption (endothermic).
Being able to calculate and interpret \(\Delta H\) helps chemists and engineers design reactions that are safe and efficient in terms of energy use.
Chemical Reactions
Chemical reactions involve the transformation of one set of substances known as reactants into another set of substances, known as products. During a chemical reaction, bonds between atoms in the reactants are broken and new bonds are formed to create the products.

In our example, the chemical reaction is the decomposition of potassium chlorate into potassium chloride and oxygen gas:
  • The balanced equation is \(2 \text{KClO}_3 (s) \rightarrow 2 \text{KCl} (s) + 3 \text{O}_2 (g)\).
A balanced equation is crucial because it ensures that the law of conservation of mass is obeyed, meaning the number of atoms of each element is the same on both sides of the equation. Understanding chemical reactions is fundamental in chemistry, as they describe how substances interact and change, impacting everything from energy production to the development of new materials.

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Most popular questions from this chapter

A piece of iron was heated to \(95.4^{\circ} \mathrm{C}\) and dropped into a constant-pressure calorimeter containing \(284 \mathrm{~g}\) of water at \(32.2^{\circ} \mathrm{C}\). The final temperature of the water and iron was \(51.9^{\circ} \mathrm{C}\). Assuming that the calorimeter itself absorbs a negligible amount of heat, what was the mass (in grams) of the piece of iron? The specific heat of iron is \(0.449 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\), and the specific heat of water is \(4.18 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\).

Colorless nitric oxide, NO, combines with oxygen to form nitrogen dioxide, \(\mathrm{NO}_{2}\), a brown gas. $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) ; \Delta H=-114 \mathrm{~kJ} $$ What is the enthalpy change per gram of nitric oxide?

Phosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4}\), can be prepared by the reactio of phosphorus(V) oxide, \(\mathrm{P}_{4} \mathrm{O}_{10}\), with water. $$ \frac{1}{4} \mathrm{P}_{4} \mathrm{O}_{10}(s)+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q) ; \Delta H=-96.2 \mathrm{~kJ} $$ What is \(\Delta H\) for the reaction involving \(1 \mathrm{~mol}\) of \(\mathrm{P}_{4} \mathrm{O}_{10}\) ? $$ \mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(a q) $$

A piece of lead of mass \(121.6 \mathrm{~g}\) was heated by an electrical coil. From the resistance of the coil, the current, and the time the current flowed, it was calculated that \(235 \mathrm{~J}\) of heat was added to the lead. The temperature of the lead rose from \(20.4^{\circ} \mathrm{C}\) to \(35.5^{\circ} \mathrm{C}\). What is the specific heat of the lead?

How does the enthalpy change for an endothermic reaction occurring at constant pressure?

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