- Calorimetry and enthalpy introduction
- Heat of formation
- Hess's law and reaction enthalpy change
- Worked example: Using Hess's law to calculate enthalpy of reaction
- Bond enthalpy and enthalpy of reaction
- Bond enthalpies
- 2015 AP Chemistry free response 7
Worked example: Using Hess's law to calculate enthalpy of reaction
Hess's law can be used to calculate enthalpy changes that are difficult to measure directly. In this video, we'll use Hess's law to calculate the enthalpy change for the formation of methane, CH₄, from solid carbon and hydrogen gas, a reaction that occurs too slowly to be measured in the laboratory. Created by Sal Khan.
Want to join the conversation?
- I am confused as to why, in the last equation, Sal takes the sum of all of the Delta-H reactions, rather than (Products - Reactants). Could someone please explain to me why this is different to the previous video on Hess's law and reaction enthalpy change.(23 votes)
- You can only use the (products - reactants) formula when you're dealing exclusively with enthalpies of formation. Here, you have reaction enthalpies, not enthalpies of formation, so cannot apply the formula.(36 votes)
- How do you know what reactant to use if there are multiple?(6 votes)
- If you are confused or get stuck about which reactant to use, try to use the equation derived in the previous video (Hess law and reaction enthalpy change). From the given data look for the equation which encompasses all reactants and products, then apply the formula. In this example it would be equation 3.(4 votes)
- If C + 2H2 --> CH4 why is the last equation for Hess's Law not ΔHr = ΔHfCH4 -ΔHfC - ΔHfH2 like in the previous videos, in which case you'd get ΔHr = (890.3) - (-393.5) - (-571.6) = 1855.4(3 votes)
- The equation for the heat of formation is the third equation, and ΔHr = ΔHfCH₄ -ΔHfC - 2ΔHfH₂ = ΔHfCH₄ - 0 – 0 = ΔHfCH₄.
890.3 is NOT the heat of formation of CH₄; it is the reverse of the combustion of CH₄.
-393.5 is NOT the heat of formation of C; it is the reverse of the combustion of C.
-571.6 is NOT the heat of formation of H₂; it is the heat of combustion of H₂
If you add all the heats in the video, you get the value of ΔHCH₄.(6 votes)
- Determine the standard enthalpy change for the formation of liquid hexane (C6H14) from solid carbon (C) and hydrogen gas (H2) from the following data :
C(s) + O2(g) → CO2(g) ΔHAo = -394.4 kJ (reaction A)
2H2(g) + O2(g) → 2H2O(l) ΔHBo = -571.6 kJ (reaction B)
2C6H14(l) + 19O2(g) → 12CO2(g) + 14H2O(l) ΔHCo = -4163.0 kJ (reaction C)
You must write your answer in kJ mol-1 (i.e kJ per mol of hexane).(4 votes)
- Why can't the enthalpy change for some reactions be measured in the laboratory?Which equipments we use to measure it?(3 votes)
- Simply because we can't always carry out the reactions in the laboratory.
For example, CO is formed by the combustion of C in a limited amount of oxygen.
But the reaction always gives a mixture of CO and CO₂.
However, we can burn C and CO completely to CO₂ in excess oxygen.
We can get the value for CO by taking the difference.(4 votes)
- where exactly did you get the other 3 equations to find the first equation?(3 votes)
- What happens if you don't have the enthalpies of Equations 1-3?(2 votes)
- Except you always do. Or you look it up in a source book.(3 votes)
- From the three equations above, how do you know which equation is to be reversed?(3 votes)
- how would we calculate the absorbed heat by sodium acetate crystal plus water while in the microwave? When the specific heat of sodium acetate and water is 0.668 and 1cal'g(2 votes)
- It seems like your question is a heat transfer problem considering your included information about the specific heat. The formula to be used is: Q = mCΔT, where Q is heat transferred, m is mass, C is specific heat, and ΔT is the change in temperature. So you need information about the mass and temperature changes of the water and sodium acetate in addition to the specific heats to be able to solve this problem.(2 votes)
I don't understand something about+ and -.
at the end of the video, when you are calculating the total ∆H.
if we know that ∆H= sum of ∆H products - sum of ∆H reactants
shouldn't it be
∆H= 890.3 -(-965.1) = 890.3 + 965 = 1185.4
instead of 890.3 + 965.1 = -74.8
thank you very much your videos are always incredibly helpful!(2 votes)
This problem is from chapter five of the Kotz, Treichel, Townsend Chemistry and Chemical Reactivity textbook. 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. 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 890.3 kilojoules per moles of the reaction going on. But if you go the other way it will need 890 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 890.3 kilojoules per mole of the reaction. 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. This is where we want to get. This is where we want to get eventually. Now, if we want to get there eventually, we need to at some point have some carbon dioxide, and we have to have at some point some water to deal with. So how can we get carbon dioxide, and how can we get water? Well, these two reactions right here-- this combustion reaction gives us carbon dioxide, this combustion reaction gives us water. So we can just rewrite those. Let me just rewrite them over here, and I will-- let me use some colors. So if I start with graphite-- carbon in graphite form-- carbon in its graphite form plus-- I already have a color for oxygen-- plus oxygen in its gaseous state, it will produce carbon dioxide in its gaseous form. It will produce carbon-- that's a different shade of green-- it will produce carbon dioxide in its gaseous form. And this reaction, so when you take the enthalpy of the carbon dioxide and from that you subtract the enthalpy of these reactants you get a negative number. Which means this had a lower enthalpy, which means energy was released. Because there's now less energy in the system right here. So this is essentially how much is released. But our change in enthalpy here, our change in enthalpy of this reaction right here, that's reaction one. I'll just rewrite it. Minus 393.5 kilojoules per mole of the reaction occurring. So the reaction occurs a mole times. This would be the amount of energy that's essentially released. This is our change in enthalpy. So if this happens, we'll get our carbon dioxide. Now we also have-- and so we would release this much energy and we'd have this product to deal with-- but we also now need our water. And this reaction right here gives us our water, the combustion of hydrogen. So we have-- and I haven't done hydrogen yet, so let me do hydrogen in a new color. That's not a new color, so let me do blue. So right here you have hydrogen gas-- I'm just rewriting that reaction-- hydrogen gas plus 1/2 O2-- pink is my color for oxygen-- 1/2 O2 gas will yield, will it give us some water. Will give us H2O, will give us some liquid water. Now, before I just write this number down, let's think about whether we have everything we need. To make this reaction occur, because this gets us to our final product, this gets us to the gaseous methane, we need a mole. Or we can even say a molecule of carbon dioxide, and this reaction gives us exactly one molecule of carbon dioxide. So that's a check. And we need two molecules of water. Now, this reaction only gives us one molecule of water. So let's multiply both sides of the equation to get two molecules of water. So this is a 2, we multiply this by 2, so this essentially just disappears. You multiply 1/2 by 2, you just get a 1 there. And then you put a 2 over here. So I just multiplied this second equation by 2. So I just multiplied-- this is becomes a 1, this becomes a 2. And if you're doing twice as much of it, because we multiplied by 2, the delta H now, the change enthalpy of the reaction, is now going to be twice this. Let's get the calculator out. It's now going to be negative 285.8 times 2. Because we just multiplied the whole reaction times 2. So negative 571.6. So it's negative 571.6 kilojoules per mole of the reaction. Now, let's see if the combination, if the sum of these reactions, actually is this reaction up here. And to do that-- actually, let me just copy and paste this top one here because that's kind of the order that we're going to go in. You don't have to, but it just makes it hopefully a little bit easier to understand. So let me just copy and paste this. Actually, I could cut and paste it. Cut and then let me paste it down here. That first one. And let's see now what's going to happen. To see whether the some of these reactions really does end up being this top reaction right here, let's see if we can cancel out reactants and products. Let's see what would happen. So this produces carbon dioxide, but then this mole, or this molecule of carbon dioxide, is then used up in this last reaction. So this produces it, this uses it. So those cancel out. Let me do it in the same color so it's in the screen. This reaction produces it, this reaction uses it. Now, this reaction right here produces the two molecules of water. And now this reaction down here-- I want to do that same color-- these two molecules of water. Now, this reaction down here uses those two molecules of water. Now, this reaction right here, it requires one molecule of molecular oxygen. This one requires another molecule of molecular oxygen. So these two combined are two molecules of molecular oxygen. So those are the reactants. And in the end, those end up as the products of this last reaction. So those, actually, they go into the system and then they leave out the system, or out of the sum of reactions unchanged. So they cancel out with each other. So we could say that and that we cancel out. And so what are we left with? What are we left with in the reaction? Well, we have some solid carbon as graphite plus two moles, or two molecules of molecular hydrogen yielding-- all we have left on the product side is some methane. So it is true that the sum of these reactions is exactly what we want. All we have left on the product side is the graphite, the solid graphite, plus the molecular hydrogen, plus the gaseous hydrogen-- do it in that color-- plus two hydrogen gas. And all we have left on the product side is the methane. All we have left is the methane in the gaseous form. So it is true that the sum of these reactions-- remember, we have to flip this reaction around and change its sign, and we have to multiply this reaction by 2 so that the sum of these becomes this reaction that we really care about. So this is the sum of these reactions. Its change in enthalpy of this reaction is going to be the sum of these right here. That is Hess's Law. So this is the fun part. So we just add up these values right here. So we have negative 393.-- no, that's not what I wanted to do. Let me just clear it. So I have negative 393.5, so that step is exothermic. And then we have minus 571.6. That is also exothermic. Those were both combustion reactions, which are, as we know, very exothermic. And we have the endothermic step, the reverse of that last combustion reaction. So plus 890.3 gives us negative 74.8. It gives us negative 74.8 kilojoules for every mole of the reaction occurring. Or if the reaction occurs, a mole time. So there you go. We figured out the change in enthalpy. And it is reasonably exothermic. Nowhere near as exothermic as these combustion reactions right here, but it is going to release energy. And we're done.