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And they say, use this information to calculate the change in enthalpy for the formation of methane from its elements. So let me just copy and paste this. Calculate delta h for the reaction 2al + 3cl2 to be. Or if the reaction occurs, a mole time. Which means this had a lower enthalpy, which means energy was released. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation. 2H2(g) + O2(g) → 2H2O(l) ΔHBo = -571. So these two combined are two molecules of molecular oxygen.
And all I did is I wrote this third equation, but I wrote it in reverse order. 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. This would be the amount of energy that's essentially released. 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. CH4 in a gaseous state. Here, you have reaction enthalpies, not enthalpies of formation, so cannot apply the formula. It's now going to be negative 285. But this one involves methane and as a reactant, not a product. Consider the reaction 2Al (g) + 3Cl(2) (g) rArr 2Al Cl(3) (g). The approximate volume of chlorine that would react with 324 g of aluminium at STP is. 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. So this actually involves methane, so let's start with this. And so what are we left with? Doubtnut is the perfect NEET and IIT JEE preparation App. Now, before I just write this number down, let's think about whether we have everything we need.
Uni home and forums. The good thing about this is I now have something that at least ends up with what we eventually want to end up with. If you add all the heats in the video, you get the value of ΔHCH₄. We figured out the change in enthalpy. 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. 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. Calculate delta h for the reaction 2al + 3cl2 x. 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. That is also exothermic.
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. To make this reaction occur, because this gets us to our final product, this gets us to the gaseous methane, we need a mole. Will give us H2O, will give us some liquid water. 6 kilojoules per mole of the reaction. It will produce carbon-- that's a different shade of green-- it will produce carbon dioxide in its gaseous form. Because i tried doing this technique with two products and it didn't work. Now, this reaction right here, it requires one molecule of molecular oxygen. 5, so that step is exothermic. So this is the fun part. Calculate delta h for the reaction 2al + 3cl2 reaction. Simply because we can't always carry out the reactions in the laboratory. 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. Further information. It has helped students get under AIR 100 in NEET & IIT JEE.
So we just add up these values right here. And when we look at all these equations over here we have the combustion of methane. I'm going from the reactants to the products. So those are the reactants.