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Coefficients, 285 balancing equations

One molecule (or mole) of propane reacts with five molecules (or moles) of oxygen to produce three molecules (or moles) or carbon dioxide and four molecules (or moles) of water. These numbers are called stoichiometric coefficients (v.) of the reaction and are shown below each reactant and product in the equation. In a stoichiometrically balanced equation, the total number of atoms of each constituent element in the reactants must be the same as that in the products. Thus, there are three atoms of C, eight atoms of H, and ten atoms of O on either side of the equation. This indicates that the compositions expressed in gram-atoms of elements remain unaltered during a chemical reaction. This is a consequence of the principle of conservation of mass applied to an isolated reactive system. It is also true that the combined mass of reactants is always equal to the combined mass of products in a chemical reaction, but the same is not generally valid for the total number of moles. To achieve equality on a molar basis, the sum of the stoichiometric coefficients for the reactants must equal the sum of v. for the products. Definitions of certain terms bearing relevance to reactive systems will follow next. [Pg.334]

Every chemical reaction can go in either forward or reverse direction. Reactants can go forward to products, and products can revert to reactants. As you may remember from your general chemistry course, the position of the resulting chemical equilibrium is expressed by an equation in which /Cec], the equilibrium constant, is equal to the product concentrations multiplied together, divided by the reactant concentrations multiplied together, with each concentration raised to the power of its coefficient in the balanced equation. Eor the generalized reaction... [Pg.152]

The coefficients of a balanced equation represent numbers of moles of reactants and products. [Pg.62]

A balanced equation remains valid if each coefficient is multiplied by the same number, including Avogadro s number, NAz... [Pg.62]

The reaction between compounds made up of A (squares), B (circles), and C (triangles) is shown pictorially below. Using smallest whole-number coefficients, write a balanced equation to represent die picture shown. [Pg.72]

Strategy (1) Start by calculating the number of moles of Fe2+. Then (2) use the coefficients of the balanced equation to find the number of moles of Mn04. Finally, (3), use molarity as a conversion factor to find the volume of KMn04 solution. [Pg.91]

The law of combining volumes, like so many relationships involving gases, is readily explained by the ideal gas law. At constant temperature and pressure, volume is directly proportional to number of moles (V = kin). It follows that for gaseous species involved in reactions, the volume ratio must be the same as the mole ratio given by the coefficients of the balanced equation. [Pg.113]

As you can see from Figure 11.1 (p. 286), the concentration of N2Os decreases with time the concentrations of N02 and 02 increase. Because these species have different coefficients in the balanced equation, their concentrations do not change at the same rate. When one mole of N2Os decomposes, two moles of N02 and one-half mole of 02 are formed. This means that... [Pg.285]

The order of a reaction must be determined experimentally it cannot be deduced from die coefficients in the balanced equation. This must be true because there is only one reaction order, but there are many different ways in which the equation for the reaction can be balanced. For example, although we wrote... [Pg.289]

If the coefficients in a balanced equation are multiplied by a factor n, rite equilibrium constant is raised to the nth power ... [Pg.327]

Note that the numbers for AP are related through the coefficients of the balanced equation (2HI, I2, H2)... [Pg.332]

Example 12.4 illustrates a principle that you will find very useful in solving equilibrium problems throughout this (and later) chapters. As a system approaches equilibrium, changes in partial pressures of reactants and products—like changes in molar amounts—are related to one another through the coefficients of the balanced equation. [Pg.333]

Express the equilibrium partial pressures of all species in terms of a single unknown, x. To do this, apply the principle mentioned earlier The changes in partial pressures of reactants and products are related through tite coefficients of the balanced equation. To keep track of these values, make an equilibrium table, like the one illustrated in Example 12.4. [Pg.335]

All the coefficients in the balanced equation are the same, 1. It follows that all the partial pressures change by the same amount, x. [Pg.336]

Given the following descriptions of reversible reactions, write a balanced equation (simplest whole-number coefficients) and the equilibrium constant expression (X) for each. [Pg.345]

The quantities E°, E x, and Eed are independent of how die equation for the cell reaction is written. You never multiply the voltage by the coefficients of the balanced equation. [Pg.489]

The coefficients a, b and c, which appear throughout these balance equations describe the extent to which these reactions occur relative to the growth reaction (ie 1 + o) and are written taken into account elemental balances for each reaction. [Pg.42]

H.4 The first box below represents the reactants for a chemical reaction and the second box the products that form if all the reactant molecules shown react. Using the key below write a balanced equation for the reaction using the smallest whole-number coefficients. Assume that if two atoms are touching,... [Pg.88]

SOLUTION We write the equilibrium constant with the partial pressure of the product, NH3, in the numerator, raised to a power equal to its coefficient in the balanced equation. We do the same for the reactants, but place their partial pressures in the denominator ... [Pg.481]

Division by the stoichiometric coefficients takes care of the stoichiometric relations between the reactants and products. There is no need to specify the species when reporting the unique average reaction rate, because the value of the rate is the same for each species. However, the unique average rate does depend on the coefficients used in the balanced equation, and so the chemical equation should be specified when reporting the unique rate. [Pg.651]

Note that the system (11.55) is valid for small deviations of the interface from Xf when mx f < 1 and exp( ix[) 1. Estimations show that the term in the thermal balance equation on the interfece is small in comparison with the term Aca3i and ALa32. Moreover, since Pg.l( g//ilg) < 1 and (/Jl/Zilg) < it is possible to neglect the second term in the expressions for coefficients an and 01.2,2 and assume that 0 31 = (ml - OgPg.l g), 32 = (ml - ol l)- Then the non-trivial solution of... [Pg.447]

There are only two possible values for concentration in a CSTR. The inlet stream has concentration and everywhere else has concentration The reaction rate will be the same throughout the vessel and is evaluated at the outlet concentration, SIa = A(ctout,bout, ) For the single reactions considered in this chapter, continues to be related to by the stoichiometric coefficient and Equation (1.13). With SS a known, the integral component balance, Equation (1.6), now gives useful information. For component A,... [Pg.22]

Rate of Formation of Primary Precursors. A steady state radical balance was used to calculate the concentration of the copolymer oligomer radicals in the aqueous phase. This balance equated the radical generation rate with the sum of the rates of radical termination and of radical entry into the particles and precursors. The calculation of the entry rate coefficients was based on the hypothesis that radical entry is governed by mass transfer through a surface film in parallel with bulk diffusion/electrostatic attraction/repulsion of an oligomer with a latex particle but in series with a limiting rate determining step (Richards, J. R. et al. J. AppI. Polv. Sci.. in press). Initiator efficiency was... [Pg.365]

The stoichiometric coefficients in a chemical equation are the smallest integers that give a balanced equation. [Pg.202]

How many moles of H2 are needed to make 4.0 moles of NH3 According to the balanced equation, 2 mol of NH3 requires 3 mol of H2. The ratio of stoichiometric coefficients from a balanced equation is called the stoichiometric ratio. We now apply this ratio to the desired amount of NH3 ... [Pg.206]

To solve a quantitative limiting reactant problem, we identify the limiting reactant by working with amounts in moles and the stoichiometric coefficients from the balanced equation. For the ammonia synthesis, if we start with 84.0 g of molecular nitrogen and 24.2 g of molecular hydrogen, what mass of ammonia can be prepared First, convert from... [Pg.219]

All changes are related by stoichiometry. Each ratio of changes in amount equals the ratio of stoichiometric coefficients in the balanced equation. In the example above, the changes in amounts for H2 and N2 are in the ratio 3 1, the same as the ratio for the coefficients of H2 and N2 in the balanced equation. [Pg.220]

Our goal is to link the mass of the product with the energy released. We must determine the number of moles of ammonia and take account of the stoichiometric coefficients in the balanced equation. A flowchart summarizes the calculations ... [Pg.377]

Greek s3Tnbol E means the sum of, and coeff refers to the stoichiometric coefficients in the balanced equation. [Pg.407]

For example, experimental studies show that the rate law for the reaction of O3 with NO2 to give N2 O5 and O2 is first order in each reactant 2 NO2 + O3 N2 O5 + O2 Experimental rate = [N02 ][03 ] Notice that for this reaction, the order of reaction with respect to NO2 is 1, whereas the stoichiometric coefficient is 2. This shows that the order of a reaction for a particular species cannot be predicted by looking at the overall balanced equation. We describe additional examples in Section 15-1. [Pg.1062]


See other pages where Coefficients, 285 balancing equations is mentioned: [Pg.722]    [Pg.2810]    [Pg.157]    [Pg.62]    [Pg.73]    [Pg.114]    [Pg.285]    [Pg.285]    [Pg.286]    [Pg.313]    [Pg.313]    [Pg.326]    [Pg.327]    [Pg.331]    [Pg.364]    [Pg.557]    [Pg.970]    [Pg.778]    [Pg.210]    [Pg.220]    [Pg.377]   


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