The ΔH for the oxidation of sulfur in the gas phase to SO₃ is –204 kJ/mol and for the oxidation of SO₂ to SO₃ is 89.5 kJ/mol. Find the enthalpy of formation of SO₂ under these conditions.

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Hey there folks, welcome back. Alright, so here we're going to be using the information from this table right here to calculate the delta age of the reaction for the reaction of ammonia and oxygen that produces nitric oxide and water vapor. So here we actually have this reaction written out double check. Make sure that it is balanced because sometimes we're given reactions that are now balanced. But after we examine this one, we can say that this is a balanced reaction. Okay, so what we need to do here is to take a look at this um equation where DELTA H. A reaction is going to equal to delta age of formation. So standard entropy of formation of all of the products minus the delta each of formations of the reactant. Now, if you take a look at the table here, obviously for NH three for ammonia, we're going to be using this value right here for N. O for nitric oxide. We're going to be using this one right here for water. So water here, we're told that it's water vapor, which means that it's an industry's form. So we're going to be using this form. The gasses form and not the liquid form? Because um water has two different envelops information depending on its phase. But you notice that you're not going to see um 02 here. And why is that? Because oh, to hear is in its natural state right, 02 is a diatonic molecule. It occurs like this naturally. So it's Delta H information is zero. So that's why you will never find it on any of the values. So now we just need to go ahead and plug in the numbers that we have here using these three values that we have on the table. Now. Why do we need a balanced equation? Because these molds right here are very important. Okay, so for example, for if we look at the products, the first one is N. O. Right? So for one mole notice that this is a killer joules per mole. So in one mole, so we have 91.3 kg per mole. Uh So for one mole of nitric oxide, um nitric oxide but we don't have one mole, we have four. So we do need to multiply that value by the amount of molds that we have for each of these compounds. So that's why we need a balanced equation. So I'm gonna go ahead and scroll down, just go ahead and um you know, take note of all of these values and how many moles of each we have. So we have four moles of N. 06 moles of water and four moles of ammonia. So let's go ahead and plug in all of those numbers into these into this equation right here. Alright, so delta each our reaction is going to equal to. Alright, so first we have the product, you know, we have four moles of it. Okay, and we're going to multiply it by its delta H information which was 91.3 and it is killed jules per mole. Alright, so we have one more product which is H20 gaseous state. And we had remember we had six right six rolls of that. Yes, six moles. So six moles multiplied by the delta atrial formation. For water and gaseous form which is negative 2 0.8 To notice that some of the Delta H. Informations are negative, some are positive. So it is very important to note that minus the delta H of the reactant. Remember 02? We are going to ignore. So there's only Ammonia and we have four moles of it in our balanced equation. So for most times the delta each of ammonia Just right there. It's -46.1. Yes. Kill jules per mole. Alright, so let's go ahead and find the answer for the for the products first. Alright, so multiply and add all of these together. You should get a negative number, it should be 1080 six or 85. and this will be in killed jules. So notice that the moles here cancel out right? Because we're multiplying it by the moles. So this will be for the products and then minus the reactant. So four times negative 46.1 is going to be obviously negative number one, Kill Jules. Alright, So -10 85.6. Double negative plus 1 84.4 is going to give us an answer of negative 901. and delta H of reactions is usually individuals per mole. So where is the small coming from? Well, it's obviously not per one mole of any of the reactant sits per one mole of that um reaction. Right. So it was going to be 90, I'm sorry, 901.2 kg joules per one mole of reaction. Or you can just say just kill a jewels and not include the most. But it is negative value here. Alright folks, That is our final answer. If you have any questions, let us know and we'll see you in the next video.

The ΔH for the oxidation of sulfur in the gas phase to SO₃ is –204 kJ/mol and for the oxidation of SO₂ to SO₃ is 89.5 kJ/mol. Find the enthalpy of formation of SO₂ under these conditions.

Verified Solution

Key Concepts

Enthalpy Change (ΔH)

Hess's Law

Enthalpy of Formation

The ΔH for the oxidation of sulfur in the gas phase to SO3 is –204 kJ/mol and for the oxidation of SO2 to SO3 is 89.5 kJ/mol. Find the enthalpy of formation of SO2 under these conditions.

Verified Solution

Key Concepts

Enthalpy Change (ΔH)

Hess's Law

Enthalpy of Formation

The ΔH for the oxidation of sulfur in the gas phase to SO₃ is –204 kJ/mol and for the oxidation of SO₂ to SO₃ is 89.5 kJ/mol. Find the enthalpy of formation of SO₂ under these conditions.