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Pdf com  questões de vestibulares sobre o tema

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Pdf com  questões de vestibulares sobre o tema

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ENTALPIA DE LIGAÇÃO - ENERGIA DE LIGAÇÃO

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🧪 VARIAÇÃO DA ENTALPIA: ENERGIA DE LIGAÇÃO

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Tabelas, fórmulas ou outras imagens que colaborem com a resolução de exercícios

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Tabelas, fórmulas ou outras imagens que colaborem com a resolução de exercícios

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https://exercicios.mundoeducacao.uol.com.br/exercicios-quimica/exercicios-sobre-energia-ligacao.htm

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Letra e).
Dados do exercício:
Ligação C-H: 105 kcal.mol ⁻¹
Ligação Cl-Cl: 58 kJ/mol
Ligação H-Cl = 103 kcal.mol ⁻¹
Ligação C-Cl = ?
Ligação C-C = 368 kcal.mol ⁻¹
ΔH = -35 Kcal.mol⁻¹.

Inicialmente é importante reescrever a equação química, demonstrando todas as ligações existentes em cada molécula:

Em seguida, devemos determinar a quantidade de energia absorvida no reagente (Er) e liberada no produto (Ep) da reação:

Para o reagente: energia absorvida (sempre positiva, independentemente do sinal dos valores fornecidos). Nos reagentes, temos o etano com a ligação C-H seis vezes e com a ligação C-C uma vez, e o reagente Cl ₂ com a ligação Cl-Cl uma vez. Assim, temos:

Para o produto: energia liberada (sempre negativa, independentemente do sinal dos valores fornecidos). Nos produtos, temos 1 mol do cloroetano, que apresenta a ligação C-H cinco vezes, a ligação C-C uma vez e a ligação C-Cl uma vez, além de 1 mol do produto HCl, que apresenta a ligação H-Cl uma vez. Assim, temos:

Com esses dados, utilizamos a expressão para o cálculo do ΔH envolvendo a energia de ligação:

ΔH = energia absorvida + energia liberada
-35 = 1056 + (-996 - x)
-35 = 1056 - 996 - x
x = 1056 - 996 +35
x = 1091 - 996
x = 95 kcal.mol ⁻¹
Resposta Questão 2
Letra a).
Dados do exercício:
Inicialmente é importante reescrever a equação química, demonstrando todas as ligações existentes em cada molécula:

Em seguida, devemos determinar a quantidade de energia absorvida no reagente (Er) e liberada no produto (Ep) da reação:

Para o reagente: energia absorvida (sempre positiva, independentemente do sinal dos valores fornecidos). Nos reagentes, temos 1 mol do CH ₄, com a ligação C-H quatro vezes, e 2 mol de O ₂, no qual temos a ligação O=O. Assim, temos:

Para o produto: energia liberada (sempre negativa, independentemente do sinal dos valores fornecidos). Nos produtos, temos 1 mol de CO ₂, que possui duas ligações C=O, e 2 mol de água, que possui duas ligações H-O. Assim, temos:

Com esses dados, basta utilizar a expressão para o cálculo do ΔH envolvendo a energia de ligação:

ΔH = energia absorvida + energia liberada
ΔH = 2656 + (-3188)
ΔH = - 532 kJ.mol ⁻¹
Resposta Questão 3
Letra b).
Dados do exercício:
Inicialmente é importante reescrever a equação química, demonstrando todas as ligações existentes em cada molécula:

Em seguida, devemos determinar a quantidade de energia absorvida no reagente (Er) e liberada no produto (Ep) da reação:

Para o reagente: energia absorvida (sempre positiva, independentemente do sinal dos valores fornecidos). Nos reagentes, temos o N ₂ com a a ligação N≡N e 3 mol de H ₂ , que apresenta a ligação H-H. Assim, temos:

Para o produto: energia liberada (sempre negativa, independentemente do sinal dos valores fornecidos). Nos produtos, temos 2 mol de NH ₃, que possui três ligações N-H. Assim, temos:

Com esses dados, basta utilizar a expressão para o cálculo do ΔH envolvendo a energia de ligação:

ΔH = energia absorvida + energia liberada
ΔH = 2252 + (-2340)
ΔH = 2252 - 2340
ΔH = - 88 kJ.mol ⁻¹

Como o enunciado pede o valor do ΔH a cada mol de NH ₃ e o valor encontrado é referente a 2, temos:

ΔH = -44 kJ.mol ⁻¹
Resposta Questão 4
Letra b).

Dados do exercício:

Inicialmente devemos determinar a quantidade de energia absorvida no reagente (Er) e liberada no produto (Ep) da reação:

Para o reagente: energia absorvida (sempre positiva, independentemente do sinal dos valores fornecidos). Nos reagentes, temos 1 mol do N ₂ com a ligação N≡N e 1 mol do O ₂ , que apresenta a ligação O=O. Assim, temos:

Para o produto: energia liberada (sempre negativa, independentemente do sinal dos valores fornecidos).

Com esses dados, basta utilizar a expressão para o cálculo do ΔH envolvendo a energia de ligação:

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exercícios resolvidos sobre energia de ligação

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questões de vestibular sobre energia de ligação.

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Exercícios sobre energia de ligação - Mundo Educação

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Letra e). Dados do exercício: Ligação C-H: 105 kcal.mol -1 Ligação Cl-Cl: 58 kJ/mol Ligação H-Cl = 103 kcal.mol -1 Ligação C-Cl = ? Ligação C-C = 368 kcal.mol -1 ΔH = -35 Kcal.mol-1.

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Tem que baixar o pdf

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Título: Design, development, and implementation of 3D-printed polylactic acid centrifuge rotors for laboratory-scale applications

Autores: Luis F. Aiquipa MorenoI; Alexander G. Guillén VasquezI; Elizabeth C. PastranaI,II,*; Roxana Y. PastranaI,#
Clique para visualizar o artigo

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Química Nova- Entalpias de Ligação Metal de Transição-Ligando em Complexos do Tipo M(

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Título: Design, development, and implementation of 3D-printed polylactic acid centrifuge rotors for laboratory-scale applications Autores: Luis F. Aiquipa MorenoI; Alexander G. Guillén VasquezI; Elizabeth C. PastranaI,II,*; Roxana Y. PastranaI,# Clique para visualizar o artigo

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- [Voiceover] We're gonna be talking about bond enthalpy and how you can use it to calculate the enthalpy of reaction. Bond enthalpy is the energy that it takes to break one mole of a bond. So one mole of a bond. So different types of bonds will have different bond enthalpies. So as an example, we can talk about a carbon hydrogen bond, or a carbon hydrogen single bond. So this carbon is probably attached to some other stuff, because carbons usually have more than one single bond. But we're gonna ignore everything else attached to the carbon, we're just gonna represent it as a big blob, like popcorn, maybe it's a protein, it could be, it could be a sugar molecule, it could be a lot of things. But we're ignoring that blob. And one other thing I forgot to say earlier is that this is the energy it takes to break one mole of a bond in the gas phase. So it's a pretty specific definition. So in the case of our carbon hydrogen bond, the bond enthalpy of this bond, so if we break this bond... Let's do a sort of dotted line. If we break this bond... We have to add energy, and what we'll get as our products is we'll get our popcorn, and what happens is, when we break this bond, the two electrons that originally made up the bond, one of the electrons will go to the carbon, and the other electron'll go to the hydrogen. And we usually represent single electrons like that using a single dot, sort of like when you write Lewis structures, you can write lone pairs with two dots. So here's our carbon with one dot, or one electron, and our hydrogen with one electron, and these are both still in the gas phase. So the delta H of this reaction is the bond enthalpy, which I will abbreviate as BE. So some important things to remember about bond enthalpy are that bond enthalpy is always positive. So it's always going to take energy, you're always gonna have to add energy to break a bond. If we take the reverse of the bond, if we take the reverse of the bond enthalpy, so another way to think about this is to flip this reaction, so if we take the reverse of this reaction, that means we're making a bond. And since we know that breaking a bond always takes energy, that means making a bond always releases energy. So it will always be negative to make a bond. And that's another way of saying, it will always release energy. And then the third thing that we're gonna discuss about bond enthalpy is that you can use it, you can use bond enthalpy to estimate delta H of reaction. And delta H of reaction is, or the enthalpy of reaction, is something that chemists are often interested in. We wanna know if it's exothermic or endothermic. You might know that there's lots of other ways of calculating delta H of reaction, such as using Hess's law... Or another way is using delta H of formation. And then there are other ways too. So this is just another way that we can use to calculate delta H of reaction using bond enthalpies. So we're gonna go through an example of that next. So the example reaction is taking propyne, which is C3H4 gas... And reacting it with hydrogen, so hydrogen gas, to get propane, C3H8 gas. And I don't know about you, I'm pretty bad at looking at a chemical formula like this and knowing exactly what the molecule looks like, so I'm gonna draw out the Lewis structures. So the Lewis structure for propyne, propyne has three carbons, and one triple, one carbon carbon triple bond, and then it has four hydrogens. So that's propyne. And we also have hydrogen gas. And our product is propane, so propane has all single bonds. So three carbons with single bonds, and eight hydrogens bound to the carbons. So that's the reaction we are interested in, and what we wanna know here is what is delta H of reaction? And how can we calculate it using bond enthalpies? We said earlier that bond enthalpies, a bond enthalpy is the energy it takes to break a bond. So what we're gonna do next is look at our reaction in terms of what bonds are broken and what bonds are formed. And this is a lot easier to do using the Lewis structures. First let's talk about which bonds are broken. We, if we compare our reactants and our products, we're breaking this carbon carbon triple bond. If we're breaking this carbon carbon triple bond, and we're also gonna break this hydrogen hydrogen bond, and one thing we forgot to do earlier which is super important, is we actually need to make sure our reaction is balanced. And we have four plus two, six hydrogens on our reactant side, and we have eight hydrogens on our product side. That's not balanced. So we actually need two hydrogen molecules on the reactant side. So let's draw one more in. So yes, we said we are breaking a hydrogen hydrogen bond, we're actually breaking two hydrogen hydrogen bonds. It's important to keep track of how many of each type of bond we're breaking because the bond enthalpy is per mole, so if you have twice as many moles it'll take twice as much energy to break all of those bonds. And then we can look at the bonds that are formed. So we have, not, since we broke this carbon carbon triple bond, that means we needed to make a new bond, and the new bond we made in our product molecule is this carbon carbon single bond. Not only did we form a new single bond between these two carbons, but now these carbons are attached to a bunch of hydrogens, so we made four new carbon hydrogen bonds. So let's write that out so that we can keep track of them when we do our final calculation of delta H of reaction. So if we just look at the bonds broken... The bonds we broke... We have a carbon carbon triple bond, and we have a couple hydrogen hydrogen bonds. Let's also just write down how many of each we have, because we'll need that for our calculation. So we have one carbon carbon triple bond, and we have two hydrogen hydrogen bonds that are broken. And then we can also look up their bond enthalpies, which are in kilojoules per mole. Bond enthalpies you can typically look up in your textbook or online, and they usually come in a table of bond enthalpies. And so the units can be kilojoules per mole, sometimes you'll also see calories or kilocalories per mole. I already looked up these bond enthalpy values. So carbon carbon triple bonds have a bond enthalpy of 835 kilojoules per mole, and hydrogen hydrogen bonds have a bond enthalpy of 800, sorry, 436 kilojoules per mole. And then next, if we look at the bonds that are broken, we have a carbon carbon single bond. And we have one of those bonds forming. And the bond enthalpy for that, which is also in terms of kilojoules per mole, is 346. And last but not least, we have the carbon hydrogen bonds that we're forming, and we have four of those, and each of those, the bond enthalpy is 413 kilojoules per mole. So now we can take all of this information and put it together to calculate delta H of reaction. So delta H of reaction, if we're thinking about it in terms of bonds made and broken, it's a total energy change during a reaction. And so it's just the energy it takes to break all of our bonds in the reactants... So to break this carbon carbon triple bond and the two hydrogen hydrogen bonds, plus the energy it takes to make the bonds, to make new product bonds. We said earlier that you always have to add energy to break bonds, so bond enthalpy is always positive, so we know this part of our calculation should always be a positive number. What that means is that it always releases energy to make new bonds, and when energy is released, delta H becomes more negative. So this number here, when we're talking about adding up the energy it takes to make new bonds, these should be negative numbers. So now let's plug in the values we have for bond enthalpy for all of these bonds that are made and broken in our reaction. Let's start with the bonds that are broken. So we have our carbon carbon single bond, that will require 835 kilojoules per mole, and we have only one of them. And we also have to break two hydrogen hydrogen bonds, so two times 436 kilojoules per mole, which is the bond enthalpy of that bond. So that's all of the bonds we break. Now we have to add up the energy that's released when we make the new bonds. So we have this carbon carbon single bond. So that is 346 kilojoules per mole, and that's negative, because that energy is released. And then, the last bond is the carbon hydrogen bond, also negative because the energy is released, and we have four of them, and each of them will release 413 kilojoules per mole. So if we still all of this into our calculator to get our final answer, what I got was that the delta H of reaction for this, for this hydrogenation reaction between propine and hydrogen gas, is -291 kilojoules per mole. We can see that this overall reaction releases energy, because delta H is negative, so it's exothermic. And that's how you can use bond enthalpies to calculate delta H of reaction.

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Bond enthalpy and enthalpy of reaction (video) | Khan Academy

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Voiceover] We're gonna be talking about bond enthalpy and how you can use it to calculate the enthalpy of reaction. Bond enthalpy is the energy that it takes to break one mole of a bond. So one mole of a bond. So different types of bonds will have different bond enthalpies.