Why Use Hess's Law in Chemical Reactions?

[mishima]

Hi, I'm trying to understand why the Hess's law process which resembles addition of equations is used as opposed to other methods, such as the sum of the products minus sum of reactants formula:

ΔH°rxn = Σ ΔH°f (products) minus Σ ΔH°f (reactants)

I get how its for instances such as rust or diamond formation where a laboratory could not replicate the process conveniently, but couldn't you also use the formula in those cases, too?

Elementary books often have questions like "verify an equation addition process with the formula", but why in practice could you only use one or the other? Wouldn't you be using all the same information either way (moles and enthalpies of formation)?

Thanks.

 

[mishrashubham]

Hi, I'm trying to understand why the Hess's law process which resembles addition of equations is used as opposed to other methods, such as the sum of the products minus sum of reactants formula:

ΔH°rxn = Σ ΔH°f (products) minus Σ ΔH°f (reactants)

I get how its for instances such as rust or diamond formation where a laboratory could not replicate the process conveniently, but couldn't you also use the formula in those cases, too?

Elementary books often have questions like "verify an equation addition process with the formula", but why in practice could you only use one or the other? Wouldn't you be using all the same information either way (moles and enthalpies of formation)?

Thanks.

I'm not sure I get your get question correctly. Could elaborate a little?

 

[mishima]

Sure, sorry. In our book, a common end of chapter problem is to find the enthalpy of reaction for a given thermochemical equation. The problem first asks you to find it by manipulating formation equations based on the reaction given. These equations would have elements in their natural state on one side, and compounds on the other. Then you look up the enthalpy of formation for that compound. You have one of these types of formation equations for each compound in the original thermochemical equation.

Next you must manipulate these intermediate equations in order to obtain the original equation. There are 2 things you can do: reverse the equation, in which case you muse change the sign of the tabulated enthalpy of formation; second you might have to multiply the entire formation equation by some coefficient, in which case you must also multiply the enthalpy of formation by the same coefficient. By doing so, the elements on the products and reactants side generally cancel out, and leave you with the original equation containing compounds.

The final step in these types of problems is to sum your modified enthalpies of formation. The sum is the enthalpy of reaction for the original equation.

Now, the next part of these problems will ask you to verify your result using the formula I gave above. If you obtain the same number, you did the Hess's Law problem correctly, is the idea (this is a high school text).

Finally, my question: why would a practicing chemist need to use the first method as opposed to using the second? The second method seems much easier to me, and my students. As I understand, the reason this process is used at all is for either very fast or very slow reactions (like graphite into diamond) which are inappropriate for methods like bomb calorimetry. I suppose in short I'm just asking if a chemist would ever use the first method above, with manipulating the formation equations, or if its just a pedagogical device.

Thanks.

 

[pa5tabear]

Sure, sorry. In our book, a common end of chapter problem is to find the enthalpy of reaction for a given thermochemical equation. The problem first asks you to find it by manipulating formation equations based on the reaction given. These equations would have elements in their natural state on one side, and compounds on the other. Then you look up the enthalpy of formation for that compound. You have one of these types of formation equations for each compound in the original thermochemical equation.

Next you must manipulate these intermediate equations in order to obtain the original equation. There are 2 things you can do: reverse the equation, in which case you muse change the sign of the tabulated enthalpy of formation; second you might have to multiply the entire formation equation by some coefficient, in which case you must also multiply the enthalpy of formation by the same coefficient. By doing so, the elements on the products and reactants side generally cancel out, and leave you with the original equation containing compounds.

The final step in these types of problems is to sum your modified enthalpies of formation. The sum is the enthalpy of reaction for the original equation.

Now, the next part of these problems will ask you to verify your result using the formula I gave above. If you obtain the same number, you did the Hess's Law problem correctly, is the idea (this is a high school text).

Finally, my question: why would a practicing chemist need to use the first method as opposed to using the second? The second method seems much easier to me, and my students. As I understand, the reason this process is used at all is for either very fast or very slow reactions (like graphite into diamond) which are inappropriate for methods like bomb calorimetry. I suppose in short I'm just asking if a chemist would ever use the first method above, with manipulating the formation equations, or if its just a pedagogical device.

Thanks.

Yes, a chemist would use the second method. Or hopefully they'd have software calculate it for them.

I'm not sure I follow completely, but it sounds like the first method is the same as the second method. It just makes you think about the transition from elemental state to compound rather than just playing with numbers.

And this process is valuable becuase it helps you calculate whether a reaction will occur for any situation. You have to calculate differently for different conditions (temp, pressure, concentrations, etc), but the method is the same. Bomb calorimetry is good, but there are often significant error sources.