Hess' Law [ENDORSED]
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Hess' Law
I don't understand why balancing one of the component equations used when solving Hess's Law you don't have to change the enthalpy. For example, in 8.55, one of the component equations is 2Al(s) + O2 (g) --> Al2O3 (g) which needs to balanced by multiplying O2 by two-thirds. Normally when we make changes to the reactions we need to change the enthalpy, which makes sense to me because you are changing the amounts of bonds broken and formed, which changes the energy needed. I don't understand why balancing doesn't also necessitate recalculating enthalpy.
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Re: Hess' Law
You are right that when we make changes to reactions we need to change the enthalpy, but that is only true if the change affects the whole reaction. In relation to your example, let me demonstrate with problem 8.57. Here, one of the equations we need to balance is H2(g)+O2(g)---->H2O(l). We are told previously that the enthalpy of combustion for H2 is -286 kJ per mole, and that will only remain true when the equation is balanced. So in this case, you would need to multiply O2 by 1/2 to balance out the equation, but this will NOT change the enthalpy because -286kJ is what the enthalpy is SUPPOSED to be for the BALANCED reaction. However, if we multiplied the entire equation by 2, making it 2H2(g)+O2(g)---->2H2O(l), the enthalpy would need therefore need to be multiplied by 2, making it -572 kJ.
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Re: Hess' Law
I am still slightly confused about how you know which reaction equation to flip the direction of when adding reactions to find the net reactions enthalpy change value. Will there ever be problems in which I need to make more than one flip of an equation to move the products and reactants to the side they belong on in the net reaction?
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Re: Hess' Law [ENDORSED]
When deciding on which equation(s) to flip, I think it's easier to know if you look at the reactions you're originally given, balance them, and then compare it to the reaction you want to end up with. Then you can manipulate the reactions given to look like the final reaction! For example, question #8.55 in the book gives you the equations 2Ba(s) + O2(g) --> 2BaO(s) and 2Al(s) + O2(g) --> Al2O3(s); they want you to find the enthalpy of the reaction 3BaO(s) + 2Al(s) --> 3Ba(s) + Al2O3(s)
After balancing the reactions given, you should end up with:
2Ba(s) + O2(g) --> 2BaO(s) and
2Al(s) + 3/2 O2(g) --> Al2O3(s)!
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[keep in mind you are trying to get 3BaO(s) + 2Al(s) --> 3Ba(s) + Al2O3(s)]
If you compare these to the reaction you want (3BaO(s) + 2Al(s) --> 3Ba(s) + Al2O3(s)), you can see that in the given reaction, Ba is on the reactants side whereas in the reaction you want, Ba is on the products side. Likewise, BaO is on the products side, and it's on the reactant side in the reaction you want to find. That's when you flip the reaction! Giving you:
2BaO(s) --> 2Ba(s) + O2(g) and
2Al(s) + 3/2 O2 (g) --> Al2O3(s)
You would not flip the second equation because the reactions and products are already on the side you want them to be in the final reaction!
For the rest of the problem, you would just manipulate the reactions to look like the final reaction by multiplying them by coefficients that would help you end up with the final reaction. Just remember that whatever you do to the reaction, you would do to delta H. Meaning, if you multiply the whole equation by a coefficient, you would multiply delta H by the same coefficient. Also, when you reverse/flip a reaction, make sure you change the sign of delta H!
After balancing the reactions given, you should end up with:
2Ba(s) + O2(g) --> 2BaO(s) and
2Al(s) + 3/2 O2(g) --> Al2O3(s)!
___________________________
[keep in mind you are trying to get 3BaO(s) + 2Al(s) --> 3Ba(s) + Al2O3(s)]
If you compare these to the reaction you want (3BaO(s) + 2Al(s) --> 3Ba(s) + Al2O3(s)), you can see that in the given reaction, Ba is on the reactants side whereas in the reaction you want, Ba is on the products side. Likewise, BaO is on the products side, and it's on the reactant side in the reaction you want to find. That's when you flip the reaction! Giving you:
2BaO(s) --> 2Ba(s) + O2(g) and
2Al(s) + 3/2 O2 (g) --> Al2O3(s)
You would not flip the second equation because the reactions and products are already on the side you want them to be in the final reaction!
For the rest of the problem, you would just manipulate the reactions to look like the final reaction by multiplying them by coefficients that would help you end up with the final reaction. Just remember that whatever you do to the reaction, you would do to delta H. Meaning, if you multiply the whole equation by a coefficient, you would multiply delta H by the same coefficient. Also, when you reverse/flip a reaction, make sure you change the sign of delta H!
Re: Hess' Law
In high school my chem teachers taught me to look at the final equation first and underline the reactants and box the products, then go to the equations given in the question and do the same that way you could see what really needs to go where. In terms of balancing, that shouldn't affect your enthalpy because they pretty much gave you the enthalpy for the balanced equation. You only change the enthalpy if you changed the equation as a whole (i.e. flipping the products and reactants, or multiplying the equation by a number).
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Re: Hess' Law
Can we treat Hess' Law problems in a similar fashion like that of an elimination/substation algebraic problems we did in high school? If so, I think that treating the molecules like constants.
What's the most efficient way to approach solving Hess' law equations?
What's the most efficient way to approach solving Hess' law equations?
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