Find ∆H° in kilojoules for the reaciton of nitric oxide with oxygen, 2 NO(g) + O2(g) → N2O4(g), given the following data: N2O4(g) → 2 NO2(g) ∆H° = 55.3 kJ NO(g) + 1/2 O2(g) → NO2(g) ∆H° = -58.1 kJ

Question

Accepted Answer

hey everyone in this example, we're given the below data and we need to find our entropy change for our final reaction here. So our first step is to recognize how our first given reaction compares to our final reaction and we should recognize that our two moles of nitrogen gas should be on the react inside of our final equation. So we're going to want to start by saying that we should reverse this reaction and that includes reversing our sign for entropy. We also should recognize that we have too many moles of all of our agents. And so we're going to go ahead and divide this entire equation by two so we can go ahead and just multiply this by one half. So what that would give us for our manipulated equation is now we have one mole of N two gas on the reactant side Added to three moles of water on our react inside as well. Then this is going to produce 3/2 moles of our oxygen gas And now two moles of our ammonia gas. We also want to go ahead and recalculate our entropy value here. And so we would take our negative 1530 kg jewels And multiply that by 1/2 And that's going to give us a value of 765 kg jewels when we reverse the sign of our entropy, which should make it positive here. So this is our first manipulated equation and now we want to go ahead and analyze our second equation. So in comparison to the final reaction, our second equation has all of the re agents on the proper sides of our equation. However, we have too many moles of all of our agents. So we're going to take these re agents and we're going to go ahead and Multiply everything by 1/2. So that would also include multiplying our entropy value by 1/2. And so now we would have a new equation where we have 5/2 moles of our oxygen gas Added to two moles of our ammonia gas, producing two moles of our nitrogen monoxide gas. And now we have three moles of our liquid water. And now our entropy value here is going to be negative 1170 kg joules times one half. And that's going to give us negative 5 85 kg joules. So we've maintained our sign here for entropy due to the fact that we haven't reversed our equation. We just we're multiplying it by one half. So that just gave us negative 5 85 kg joules. And now our next step is to go ahead and add these two equations together to see how they give us our final entropy value. And if they match up to our final equation, so we should recognize that we can cancel out our three moles of water because we have them in both equations. We also can go ahead and get rid of the two moles of our Ammonia gas in both equations. And so this is going to give us our final equation where we have one mole of co two gas Added to one mole of nitrogen gas, producing two moles of nitrogen monoxide gas. And when we take the two entropy values here and add them up together, this is positive 7 65 plus negative 5 85 kg joules, Which is going to give us a final entropy value of 180 kg jewels. And so this is going to be our final answer as our change in entropy for our final given reaction here. So I hope that everything we went through was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

Find ∆H° in kilojoules for the reaciton of nitric oxide with oxygen, 2 NO(g) + O2(g) → N2O4(g), given the following data: N2O4(g) → 2 NO2(g) ∆H° = 55.3 kJ NO(g) + 1/2 O2(g) → NO2(g) ∆H° = -58.1 kJ

Verified Solution

Key Concepts

Hess's Law

Standard Enthalpy of Formation (∆H°f)

Enthalpy Change (∆H)