5 answers

AH? of AlOj(s) (hint: first write the chemical 29. Using the following thermochemical data, calculate for AH" of AlzO; and

Question:

AH? of AlOj(s) (hint: first write the chemical 29. Using the following thermochemical data, calculate for AH" of AlzO; and then use Hess Law reaction 6HCIg) 4H 388_ kJlmol 2AICI;(s) 3HzO() AlzO3(s) AH -1973.2 kJlmol ZAI(s) 3Clzkg) 2AICI;(s) AH 202.4 kJmol 4HClg) + Oz(g) ~> 2Chz(g) 2HzO() Answer: -1888 kJlmol


Answers

Use the information in thermochemical equations (1) through ( 3 ) to calculate the value of $\Delta H_{\mathrm{rxn}}^{\circ}$ for the reaction in equation (4). (1) $\mathrm{Pb}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{PbO}(s) \quad \quad \Delta H_{\mathrm{rxn}}^{\circ}=-219 \mathrm{kJ}$ (2) $\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) \Delta H_{\text {rxn }}^{\circ}=-394 \mathrm{kJ}$ (3) $\mathrm{PbCO}_{3}(s) \rightarrow \mathrm{PbO}(s)+\mathrm{CO}_{2}(g) \quad \Delta H_{\text {rxn }}^{\circ}=86 \mathrm{kJ}$ (4) $2 \mathrm{Pb}(s)+2 \mathrm{C}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{PbCO}_{3}(s) \quad \Delta H_{\mathrm{rxn}}^{\circ}=?$

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This problem will be using Hester's law to determine the entropy of the reaction given a bunch of other reactions. So you were actually born, even for is that wanted lead? And we're gonna be using be on three to determine that. So let's start by looking at it they need to see during h four, you said it's on the left of the reaction Washington before, So it's also in the second equation, and it's also on the left. So where is going to rewrite that equation? And the entropy is going to stay the same since we're not flipping it or multiplying by anything. And I just left off the States for all the gases.

Water is the only once in liquid form just for simplification. So this is just gonna be negative 1937 cordials for more. And then let's look at the product side on the reaction we're looking for. We have that C three h eight that of years in the third reaction, but this time it's on the left. So we're gonna do is foot that reaction when we rewrite it and we're also gonna have to flip the sign on the implication of that reaction.

So it's gonna be cause again. 1000 You under 19 1 Phil jewels normal. And then we can add those reactions together. We're going to cancel out the CEO, choose we're gonna be left with one oxygen on the right and two waters on the left. That's going to give us this reaction.

And that's still not the reaction we're looking for. So we're gonna have to use that formation of water reaction to get to the reaction we're shooting for. So we have to water molecules on the left so we can it multiply that formation of water reaction by two so we can get those waters to cancel out. So we're gonna have to register anymore on oxygen and two waters and that until you is gonna be multiplied by two since we multiply it all of the coefficients by two opposing me two times that negative to 85 0.8 Jules for more. Now, if we had used to reactions together, cancel out the oxygen water and it's gonna leave us with the reaction you're looking for.

And finally, we can just add these numbers up here and I will give us the Delta H reaction. How long are you left? With negative to a nine points and that's money and kill jewels. More worry. The end will be about reaction..

So there is a mistake in this question for that third reaction that say that the Delta h of the reaction is negative. Um, 100 90.4 killer jewels per mole. That would be if the chemical reaction were flipped and methane was reacting with oxygen forming carbon dioxide and water. So the Delta H for the chemical reaction as written is actually positive. 890 killed Jules for mold.

So the first thing to do then is write the chemical reaction that describes what we desire the Delta H of formation of methane. So one mole of carbon reacts with two moles of hydrogen get gas to produce one mole of methane. Then the chemical reactions that are given to us are shown here below, and I'll just rewrite them so that we have them before way begin manipulating and using has a lot to get belt h of the reaction about. So you'll notice then that I have read written this here as positive. This is the only way you're able to get the answer in the back of the book.

So the first thing that we need to do is multiply the second reaction by two. In doing so, we'll get to hydrogen is on the left hand side, which is what we need. And that's all we need to dio. You will notice that, um, the to oxygen's two moles of Oxygen's Here Castle, the two moles of oxygen's here. The two moles of liquid water canceled two moles of liquid water here and carbon dioxide dear will cancel the carbon dioxide here we can, then some up to chemical reaction in order to get the chemical reaction that we desire and then some about all the modified Delta H values.

And we get negative 74.9 killer jewels..

Hello Today we'll be talking about Chapter 15 Question 94 which asks us to consider the net reaction generically shown at the top to a plus B to C three goes to two B plus A to C three. Now we need to calculate Delta H for this reaction using the two sub reactions shown A and B. So let's look att equation a first to a plus three have C two goes to A to C three, and the important thing to note is that we have to A and A to C three, and they match up well with the net equation there on the right side of the equation, the correct sides of the equation. So we don't need to do anything to equation a equation. Be to be plus three have C two goes to B to C three also has B to C three and to be, but now they're on the incorrect sites there on the flip sides of the equation from what we need.

So let's just rewrite this equation to be to have you be to see three and be on the correct side of the equation that we want. B to C three goes to to be plus three halves See, too, because we reverse the direction of this reaction. We just have to put a negative sign in front of the Delta H reaction. Negative to 85 Killer Jules now, And I'm gonna cross out the middle line now. So now we have to be to see three on the left side of the equation here and in the net reaction and be on the right side.

This means that we can combine sub equations A and B the reverse version of be here and hopefully some them to get the net reaction and also to get the Delta H reaction. And if we look, we have three have C two after the arrow on the bottom equation, and three have C two before the arrow on the top equation, which means we can cancel both of them out. And that leaves us with the net equation to a plus B to C three shown here goes to to be plus A to C three. We've generated the net reaction by just adding these two equations. And so because that worked, we can just add the delta H reactions as well.

And so we just have negative. 1874. Kill it, Jules. Plus negative 285 Killer Jewel's to give us the Delta H for the total reaction, which is equal to negative 2159. Killer Jules.

And so we can see that we can take Delta H is of known sub reactions and just combine them by reorganizing them, reversing the directions and changing the Delta h reaction slightly and then some them all together to get the Delta age for some net reaction. Thank you..

Here we are given a series of reactions reaction 12 and three. And what we'd like to do is calculate the entropy of the formation of one mole of die nitrogen pens oxide. So firstly, what we'd like to do in step is reverse reaction one Lord supplied by 1/2. Give us reaction for Delta H is equal to positive 285.8 kg joules than a poppy. Reverse reaction to To obtain reaction five Delta H is equal to positive 73.7 kg jewels In part c.

Multiply reactions three by a factor of two. To obtain reaction six, adult H is equal to negative 348.2 kg jewels, continuing on to part D. Here. So we add reaction five and 6 together located here and here. Give us reaction seven but don't age is equal to negative 274.5 kg jewels.

So then we combine Reaction one and 7 To obtain the reaction for the formation of dying nitrogen oxide. Where Delta H is equal to positive 11.3 kg jewels..

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