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Why are fractional coefficients permitted in a balanced thermochemical equation? If a formula in a thermochemical equation has a coefficient of \(\frac{1}{2}\), what does it signify?

Short Answer

Expert verified
Fractional coefficients are allowed to ensure the conservation of mass is maintained. A coefficient of \(\frac{1}{2}\) signifies that only half a molecule or mole of the reactant or product takes part in the chemical reaction.

Step by step solution

01

Understanding the Role of Fractional Coefficients

Fractional coefficients in balanced thermochemical equations are permitted because they indicate that the reaction involves a fractional part of a reactant or product molecule. This is often necessary when ensuring that the equation obeys the law of conservation of mass, wherein the number of atoms of each element must be the same on both sides of the equation.

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Key Concepts

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

Fractional Coefficients in Chemistry
Fractional coefficients in chemical equations are introduced to maintain the strict rule of the conservation of mass. When balancing chemical equations, we must ensure that the same number of atoms of each element is present on both sides of the equation. At times, to achieve this balance, we use fractions to represent the quantities of reactants or products involved in the reaction.

Fractional coefficients, while mathematically unconventional, are scientifically significant. They indicate that only a part of a molecule is involved in the reaction. For example, a coefficient of \(\frac{1}{2}\) before oxygen, as in \(O_2\), means that the reaction only uses half a molecule of oxygen, which in practical terms would be interacting with another half from some other part of the reaction or another reaction altogether, to form a complete molecule.

It's important to recognize that in a laboratory setting, reactions don't occur with half molecules; these coefficients are used for stoichiometric calculations to accurately represent proportional relationships. When scaling up the reaction to a number that gives whole numbers for all reactants and products, the fraction can be eliminated, ensuring practical application and ease of measurement.
Law of Conservation of Mass
The law of conservation of mass is a fundamental principle stating that mass is neither created nor destroyed in a chemical reaction. Consequently, the mass of the reactants must equal the mass of the products. This law is the cornerstone of chemical equation balancing.

When a chemical equation is written, it must reflect this principle by showing the same number of atoms for each element on both sides. Fractional coefficients may sometimes be the only means to achieve this balance without violating the law. Understanding this law helps explain why fractional coefficients are not only acceptable but sometimes necessary in representing certain chemical reactions.
The practical implication of this law is that in any chemical process, including reactions in industrial manufacturing or in metabolic processes in the body, both sides of the equation must be accounted for, ensuring material balance and enabling prediction of yield and reactant consumption.
Balancing Chemical Equations
Balancing chemical equations is a critical skills in chemistry that ensures the law of conservation of mass is respected. It involves adjusting the coefficients of reactants and products so the number of atoms for each element is the same on both sides of the equation.

To balance an equation, start by listing the number of atoms of each element in the reactants and products. Then, adjust the coefficients—one at a time—until the equation is balanced. Remember, coefficients multiply the entire molecule, including all atoms present in the molecule. If you end up with fractional coefficients, try to multiply all coefficients by the smallest number necessary to obtain whole number coefficients, simplifying the equation while keeping it balanced.

Consistent practice in balancing equations not only enhances understanding of chemical reactions but also aids in the development of problem-solving skills pertinent to various scientific enquiries and practical applications, such as calculating reagent needs or analyzing reaction yields.

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Most popular questions from this chapter

For samples with the same mass, which kind of substance needs more energy to undergo an increase of \(5^{\circ} \mathrm{C},\) something with a large specific heat or something with a small specific heat? Explain.

What term is used to describe a reaction that absorbs heat from the surroundings? How does the chemical energy change during such a reaction? What is the algebraic sign of \(q\) for such a reaction?

What two conditions must be met by a thermochemical equation so that its standard enthalpy change can be given the symbol \(\Delta H_{\mathrm{f}}^{\circ}\) ?

How are heat and work defined?

Chlorofluoromethanes (CFMs) are carbon compounds of chlorine and fluorine and are also known as Freons. Examples are Freon-11 \(\left(\mathrm{CFCl}_{3}\right)\) and Freon-12 \(\left(\mathrm{CF}_{2} \mathrm{Cl}_{2}\right),\) which were used as aerosol propellants. Freons have also been used in refrigeration and air-conditioning systems. In 1995 Mario Molina, F. Sherwood Rowland, and Paul Crutzen were awarded the Nobel Prize mainly for demonstrating how these and other CFMs contribute to the "ozone hole" that develops at the end of the Antarctic winter. In other parts of the world, reactions such as those shown below occur in the upper atmosphere where ozone protects the earth's inhabitants from harmful ultraviolet radiation. In the stratosphere CFMs absorb high-energy radiation from the sun and split off chlorine atoms that hasten the decomposition of ozone, \(\mathrm{O}_{3}\). Possible reactions are $$ \begin{array}{ll} \mathrm{O}_{3}(g)+\mathrm{Cl}(g) \longrightarrow \mathrm{O}_{2}(g)+\mathrm{ClO}(g) & \Delta H^{\circ}=-126 \mathrm{~kJ} \\ \mathrm{ClO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) & \Delta H^{\circ}=-268 \mathrm{~kJ} \\ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g) & \Delta H^{\circ}=? \end{array} $$ The \(\mathrm{O}\) atoms in the second equation come from the breaking apart of \(\mathrm{O}_{2}\) molecules caused by ultraviolet radiation from the sun. Use the first two equations to calculate the value of \(\Delta H^{\circ}\) (in kilojoules) for the last equation, the net reaction for the removal of \(\mathrm{O}_{3}\) from the atmosphere.

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