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. So they're giving us the enthalpy changes for these combustion reactions-- combustion of carbon, combustion of hydrogen, combustion of methane. Because we just multiplied the whole reaction times 2. So it's positive 890. Worked example: Using Hess's law to calculate enthalpy of reaction (video. From the given data look for the equation which encompasses all reactants and products, then apply the formula. So this is a 2, we multiply this by 2, so this essentially just disappears. I'll just rewrite it.
The good thing about this is I now have something that at least ends up with what we eventually want to end up with. Actually, I could cut and paste it. So we can just rewrite those. What happens if you don't have the enthalpies of Equations 1-3? You do basically the same thing: multiply the equations to try to cancel out compounds from both sides until youre left with both products on the right side. It gives us negative 74. Calculate delta h for the reaction 2al + 3cl2 3. 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. So how can we get carbon dioxide, and how can we get water? 8 kilojoules for every mole of the reaction occurring. And in the end, those end up as the products of this last reaction. 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. Why does Sal just add them?
A-level home and forums. That is also exothermic. In this example it would be equation 3. So I just multiplied this second equation by 2. So it is true that the sum of these reactions is exactly what we want. Calculate delta h for the reaction 2al + 3cl2 x. 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. NCERT solutions for CBSE and other state boards is a key requirement for students. 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. The equation for the heat of formation is the third equation, and ΔHr = ΔHfCH₄ -ΔHfC - 2ΔHfH₂ = ΔHfCH₄ - 0 – 0 = ΔHfCH₄.
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. You must write your answer in kJ mol-1 (i. e kJ per mol of hexane). So two oxygens-- and that's in its gaseous state-- plus a gaseous methane. 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. Calculate delta h for the reaction 2al + 3cl2 will. So these two combined are two molecules of molecular oxygen. 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. Popular study forums. Well, these two reactions right here-- this combustion reaction gives us carbon dioxide, this combustion reaction gives us water. And then you put a 2 over here. You use the enthalpy changes from a bunch of different reactions to find the enthalpy change of one reaction through eliminating other terms like he did in this video.
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. And when we look at all these equations over here we have the combustion of methane. So we could say that and that we cancel out. So this is essentially how much is released.
If you add all the heats in the video, you get the value of ΔHCH₄. Simply because we can't always carry out the reactions in the laboratory. Its change in enthalpy of this reaction is going to be the sum of these right here. So I like to start with the end product, which is methane in a gaseous form. 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). Which means this had a lower enthalpy, which means energy was released. Now, this reaction down here uses those two molecules of water. How do you know what reactant to use if there are multiple? And they say, use this information to calculate the change in enthalpy for the formation of methane from its elements. So this is the sum of these reactions. So those, actually, they go into the system and then they leave out the system, or out of the sum of reactions unchanged.
Created by Sal Khan. And we have the endothermic step, the reverse of that last combustion reaction. Or we can even say a molecule of carbon dioxide, and this reaction gives us exactly one molecule of carbon dioxide. And now this reaction down here-- I want to do that same color-- these two molecules of water. Uni home and forums. Let me do it in the same color so it's in the screen. So this produces carbon dioxide, but then this mole, or this molecule of carbon dioxide, is then used up in this last reaction. 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. So this produces it, this uses it. This would be the amount of energy that's essentially released.
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. CH4 in a gaseous state. This problem is from chapter five of the Kotz, Treichel, Townsend Chemistry and Chemical Reactivity textbook. And all we have left on the product side is the methane. 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 cancel out with each other.