Under certain nonstandard conditions, oxidation by O2( g) of 1 mol of SO2( g) to SO3( g) absorbs 89.5 kJ. The enthalpy of formation of SO3( g) is -204.2 kJ under these conditions. Find the enthalpy of formation of SO2( g).

Video transcript

Welcome back, everyone. Under certain non-standard conditions, the decomposition of sulfur trioxide into one half of the mole of oxygen gas and one mole of sulfur dioxide gas absorbs 227.8 kilojoules. The entropy of formation of sulfur trioxide gas is negative 204.2 kilojoules under these conditions, find the entropy of formation of sulfur dioxide gas. And we're given four answer choices. A negative 64.9 B 32.4 C negative 40.4 D 23.6. All of them are given in kilojoules per mole. So first of all, let's write down the reaction we have. So three gas decomposes as the problem suggests into one half of a mole of oxygen gas and a one mole of sulfur dioxide gas. We can essentially write down that the entropy change of this reaction is equal to the sum of the entropies of formation of products minus the sum of the entropies of formation of reactants. It might sound confusing but essentially what we're going to do is just write down the equation. We have one half of a mole of oxygen. So we're taking one half of a mole multiplied by the entropy of formation of oxygen gas plus. Now our next product is one mole of so two. So we take one mole multiplied by the entropy of formation of so two gas and we are subtracting the ample piece of formation of the reactants. So the only reactant is so two, we take one mole of it. It's actually so three, right? So we're taking the entropy affirmation of so three gas. Now, what do we know? Well, essentially we know that the reaction absorbs heat. What does that mean? Well, essentially we know that absorption of heat represents an endothermic reaction. So we know that the entropy change of the reaction would be equal to 227.8 kilojoules. We also know that the entropy of formation of oxygen, right. In this case, it would be zero because it is in a standard state auction is gas here. So we're going to see that it is equal to zero. What we're seeing from here is that we end up with the entropies of formation of so two and so three, what are we looking for? Well, we're looking for the entropy of formation of so two. So if we rearrange the equation, we can say that the entropy of formation of two gas would be equal to the entropy of the reaction minus the entropy of formation of. So three gas, we take 227.8 kilojoules. And now we need to use the thermodynamic tables to identify the entropy affirmation of. So three, right, according to the tables, we can essentially state that based on the fact that the entropy formation of so three gas is equal to negative 204 12 killer jules. For more, we can use this number in our equation. There's one more thing to consider actually, if we look at our equation and if we are solving for the entropy of formation of so two, that actually means that we must add the entropy of formation of so three, right? Because it was negative, we are moving it to the left. So my apologies here, we must have a positive sign. So we're taking 227.8 kilojoules and we're adding one mole multiplied by negative 204.2 kilojoules per mole. And that would be our calculation. Of course, it would be more accurate to say that here we're just adding one mole since this is exactly what we stated previously. So let's add that one mole for complete accuracy. And now let's evaluate the result. Well, essentially we end up with 23.6 killer jewels, right? We don't have any units per M. But if we're interested in the molar entropy change formation, we can essentially add and state that the entropy of formation based on this reaction because we already had one mole, we can add that unit. So the entropy formation of so two gas would be equal to 23.6 kilojoules per mole of. So two, you know, looking at the answer choices, we can state that the correct answer is option D. Thank you for watching.