Related questions How is Hess's law applied in calculating enthalpy? How do you compute Hess's law calculations? How do you draw a Hess's law diagram? What does Hess's law say about the enthalpy of a reaction? Why isn't Hess's law helpful to calculate the heat of reaction involved in converting a diamond Google Classroom Facebook Twitter.
So they tell us, suppose you want to know the enthalpy change-- so the change in total energy-- for the formation of methane, CH4, from solid carbon as a graphite-- that's right there-- and hydrogen gas. So we want to figure out the enthalpy change of this reaction.
How do we get methane-- how much energy is absorbed or released when methane is formed from the reaction of-- solid carbon as graphite and hydrogen gas? So they tell us the enthalpy change for this reaction cannot to be measured in the laboratory because the reaction is very slow. So normally, if you could measure it you would have this reaction happening and you'd kind of see how much heat, or what's the temperature change, of the surrounding solution.
Maybe this is happening so slow that it's very hard to measure that temperature change, or you can't do it in any meaningful way.
We can, however, measure enthalpy changes for the combustion of carbon, hydrogen, and methane. So they're giving us the enthalpy changes for these combustion reactions-- combustion of carbon, combustion of hydrogen, combustion of methane. And they say, use this information to calculate the change in enthalpy for the formation of methane from its elements.
So any time you see this kind of situation where they're giving you the enthalpies for a bunch of reactions and they say, hey, we don't know the enthalpy for some other reaction, and that other reaction seems to be made up of similar things, your brain should immediately say, hey, maybe this is a Hess's Law problem.
Hess's Law. And all Hess's Law says is that if a reaction is the sum of two or more other reactions, then the change in enthalpy of this reaction is going to be the sum of the change in enthalpies of those reactions. Now, when we look at this, and this tends to be the confusing part, how can you construct this reaction out of these reactions over here?
And what I like to do is just start with the end product. So I like to start with the end product, which is methane in a gaseous form. And when we look at all these equations over here we have the combustion of methane. So this actually involves methane, so let's start with this.
But this one involves methane and as a reactant, not a product. But what we can do is just flip this arrow and write it as methane as a product. So if we just write this reaction, we flip it.
So now we have carbon dioxide gas-- let me write it down here-- carbon dioxide gas plus-- I'll do this in another color-- plus two waters-- if we're thinking of these as moles, or two molecules of water, you could even say-- two molecules of water in its liquid state.
That can, I guess you can say, this would not happen spontaneously because it would require energy. But if we just put this in the reverse direction, if you go in this direction you're going to get two waters-- or two oxygens, I should say-- I'll do that in this pink color. So two oxygens-- and that's in its gaseous state-- plus a gaseous methane. CH4 in a gaseous state. And all I did is I wrote this third equation, but I wrote it in reverse order. I'm going from the reactants to the products.
When you go from the products to the reactants it will release But if you go the other way it will need kilojoules.
So the delta H here-- I'll do this in the neutral color-- so the delta H of this reaction right here is going to be the reverse of this. So it's positive All I did is I reversed the order of this reaction right there.
The good thing about this is I now have something that at least ends up with what we eventually want to end up with. Sometimes, you will need to multiply a given reaction intermediate through by an integer. Restating the first equation and flipping the second equation, we have:.
Adding these equations together, carbon dioxides and oxygens cancel, leaving us only with our net equation. Boundless vets and curates high-quality, openly licensed content from around the Internet. Of considerable importance is the observation that the heat input in equation [2], This is not a coincidence: if we take the combustion of carbon and add to it the reverse of the combustion of hydrogen, we get.
Thus, taking the combustion of carbon and "subtracting" the combustion of hydrogen or more accurately, adding the reverse of the combustion of hydrogen yields equation [2]. And, the heat of the combustion of carbon minus the heat of the combustion of hydrogen equals the heat of equation [2]. By studying many chemical reactions in this way, we discover that this result, known as Hess's Law, is general.
A pictorial view of Hess's Law as applied to the heat of equation [2] is illustrative. The reaction arrow connecting these boxes is labeled with the heat of this reaction. Now we take these same materials and place them in a third box containing C s , O 2 g , and 2 H 2 g.
This box is connected to the reactant and product boxes with reaction arrows, labeled by the heats of reaction in equation [3] and equation [4]. A consequence of our observation of Hess's Law is therefore that the net heat evolved or absorbed during a reaction is independent of the path connecting the reactant to product this statement is again subject to our restriction that all reactions in the alternative path must occur under constant pressure conditions.
A slightly different view of figure 1 results from beginning at the reactant box and following a complete circuit through the other boxes leading back to the reactant box, summing the net heats of reaction as we go. We discover that the net heat transferred again provided that all reactions occur under constant pressure is exactly zero.
This is a statement of the conservation of energy: the energy in the reactant state does not depend upon the processes which produced that state. Therefore, we cannot extract any energy from the reactants by a process which simply recreates the reactants.
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