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When 21.5 mL of 0.500 M H2SO4 is added to 21.5 mL of 1.00 M KOHin a coffee-cup calorimeter at 23.50°C, the temperature rises to30.17°C. Calculate ΔH of th...

Question

When 21.5 mL of 0.500 M H2SO4 is added to 21.5 mL of 1.00 M KOHin a coffee-cup calorimeter at 23.50°C, the temperature rises to30.17°C. Calculate ΔH of this reaction. (Assume that the totalvolume is the sum of the individual volumes and that the densityand specific heat capacity of the solution are the same as for purewater.) (d for water = 1.00 g/mL; c for water = 4.184J/g·°C.)

When 21.5 mL of 0.500 M H2SO4 is added to 21.5 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH of this reaction. (Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water.) (d for water = 1.00 g/mL; c for water = 4.184 J/g·°C.)



Answers

When $25.0 \mathrm{~mL}$ of $0.500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ is added to $25.0 \mathrm{~mL}$ of 1.00 $M$ KOH in a coffee-cup calorimeter at $23.50^{\circ} \mathrm{C}$, the temperature rises to $30.17^{\circ} \mathrm{C}$. Calculate $\Delta H$ in $\mathrm{kJ}$ per mole of $\mathrm{H}_{2} \mathrm{O}$ formed. (Assume that the total volume is the sum of the volumes and that the density and specific heat capacity of the solution are the same as for water.)

So in this neutralization reaction we're going to react some potassium hydroxide and sulfuric acid. So why don't we go ahead and start with a balanced equation? Okay, so the K. O. H. Is going to react with the sulfuric acid, H. two. So 4 And we're going to make K. two S. 04 in water. So we'll go ahead and balance those. There's two here and two here. Um And we know that the heat for this is -56 killer jewels per mole of K. O. H. Okay, so not necessarily for the equation has written but per one mole of K. O. H. So we're gonna go ahead and do a little study geometry here. So let's start with the information they gave us about sulfuric acid because we have a polarity and a volume. So if you have a polarity and a volume you can calculate moles. So So .475 moller of H two S. 04 Times 25 ml, or .025 L. So we know that .0119 moles of H two S. 04 and then we'll go ahead and do our mole ratio And we'll change moles of H. two so 4 two malls of K. A. wage And that's a 2-1 ratio. So we're going to get 0.2 38 malls of K. O. H. And we want to know the volume that was added. So we're just going to go ahead and divide this by its polarity which is moles per liter 2.613 More. And that tells us that we had we needed 0.3 87 liters or 38 0.7 mL. Mhm. So I'll see if we can keep track of the heat and how much that's going to change our temperature. Okay. So we know that um 56 killer jewels were released by one mall. Okay. A wage. But we know that we have point oh two 3 8 malls of care wage. So we're going to get 1.32 Killer Jewels released here. Okay this is a negative so this is a negative or it's released. I'm gonna go ahead and change that to jules. So that's 1320 jewels. Okay it's still negative. So now let's go ahead and plug into Q. Is negative. M. C. Delta T. So the heat being given off There's negative 1320 jewels. I changed it to jules because that's what our specific heat capacity is going to be in And now I need the mass. Well we're going to do a little work here. So we took 25 ml okay. Of the H. Two S. 04 and we added 38 0.7 mL over K. Which So we ended up with a total of 63 .7 ml of solution. And they said to assume the density is one, So we'll have 63 .7 g of solutions. So I'll go ahead and put that in here, 63.7 g. Been multiplied by our specific key capacity and jewels. Program degrees Celsius. And we're going to look for our change in temperature. So change in temperature Is going to come out to be 4.96°C.. And since we started at 23/7 hadn't add that. And we end up at 28 .7°C as our final temperature.

Hey, everybody. So in this question we have a neutralization reaction between sulfuric acid and potassium hydroxide forming assault and water. So we're giving a few known, so I'll go ahead and write those down. We know that we have 25 mills of 0.5 Mueller sulfuric acid as well as 25 mills of potassium hydroxide, which is one Mueller. We also know that the starting temperature for this reaction, called a T subzero, is equal to 23.5 degrees Celsius, and the final temperature after the reaction is over is 30.17 degrees Celsius. Now the question asks us to find the change in entropy. So that is our Delta H. With these knowns and unknowns, we can go ahead and start solving for our entropy change. So to do that, the first thing I'm going to do is convert are starting materials into their respective Moeller values. And I'm gonna use these Moeller values to find the which one of them, which one of the reactant is the limiting reacted for the reaction when forming water. So for sulfuric acid, we're starting off with 25 mills and then we're going to convert that into leaders with our conversion factor and then using the U concentration that's given to us. So for one leader, there is to your 0.5 moles of sulfuric acid that gives us a 0.125 moles for sulfuric acid now going on to potassium hydroxide. We also start with 25. Mills will convert that to leaders. So there are 1000 middle leaders and one leader and were given a concentration of one mole per leader and that yields a 0.0 25 moles for potassium hydroxide. Now that we know that we can go ahead and use these initial moral values to find how much water will be produced to try to determine the limiting reactant. So go ahead and do that. We have zero point 0125 moles of sulfuric acid and using our balanced equation from above. We know that for every one mole, both of surface sulfuric acid, it will form two moles of water. Working this out in the calculator, we find that 0.25 moles of water will form using that amount of sulfuric acid now doing the same thing, but with the potassium hydroxide. So before we calculated 0.25 moles of potassium hydroxide and then using the balanced equation, we know for every one mole of potassium hydroxide, this will form one more of water. And I went ahead and simplified the fraction since in the equation it was to over two, and I just simplified that already 2 1/1 since that's how the ratio worked out. So that also yields a zero point 0 to 5 moles of H 20 Since both of them, you did the same amount of water. We can say that both of these will yield 0.25 mils of H 20 And we can use this value when we're calculating our heat. So moving on, we can go ahead and use our heat equation to calculate the heat absorbed by the calorie bitter. To do that, we'll use our equation. Q. For the calorie emitter, is he good? EMC Delta T where m is our mass see is the heat capacity and deep delta T is the change in temperature. We have all of those values. So for the mass, I'm going to use the total volume. So we started off 25 mils of both of each reactant. So that gives me a total of 50 mills of water produced. And I'm gonna go ahead and multiply this by the density of water. So that is one gram for one mill leader and that were you in my mass from the textbook. We know that the heat capacity is 4.184 jewels programs, degrees Celsius and the change of temperature. I'll go ahead and run in the second line since I'm running out of room is the final minus the initial. So we have the final temperature 30.17 degrees, and we're going to subtract that from 23.5, plugging all of this into the calculator, we find that the Q absorbed by the calorie motor is one 1003 195 jewels, and this is the same thing as saying 1.395 killer jewels. Now that we know that the heat from the calorie meter is this, we can say that the heat for the reaction is the negative of the heat absorbed from the calorie meter. Since the reaction is releasing heat. So this is negative 1.395 killer jewels And using this number, we can finally find our change in Anthill P. So don't h is equal to our heat that we just calculated negative 1.395 killer jewels. And we're going to divide that by the moles that we calculated above for water. So that is 0.25 moles of water. Plugging that into the calculator, we find that the heat of the entropy change. Excuse me, is negative. 55 0.8 killer jewels per mol H 20 and that is our final answer.

Okay for this particular question, the first thing that we need to do is calculate the heat that is going to be released into the 150 milliliters of water when five grams of sodium hydroxide and 4.2 grams of potassium hydroxide are added. Do this. We can take the massive of sodium hydroxide converted to moles and then use the Moller Date Delta H of solution for sodium hydroxide. This will then correspond to the heat. Associate it with the addition of sodium hydroxide will add that to the heat associated with the addition of potassium hydroxide doing a similar calculation, converting its grams to moles. And then it's molds to kill the jewels. Negative 9.73 Killer jewels are released when these processes or when these substances air added toe water. So that is, the energy added the water to calculate the amount of heat that is, um, transferred to the water that is the heat transfer to the water. Figure out change in temperature that occurs when we add that amount of heat water. You recognize that the heat is transferred. Uh, the heat released is equal to the negative of the heat gained. So we're changing the heat releases negative. It's equal to the negative of the heat gained, and what I'm about to calculate is the heat gain. But I need that negative sign there so that negative signs will cancel my temperature change. So I have 150 grams. I could revert that two moles of water and use the Moller hit capacity of water that was given. Or you could use the specific it of water. Either way, you're going to get the correct answer and then multiply that by the change in temperature. However, one thing that they do need to recognize is this is in units of killer jewels, and this is in units of jewels. So I need to make sure that the heat units are the same for all convert on this right hand side theme jewels that I calculate to kill a jewels. Then I can do the rest of the algebra and solve for T Final, which is 38.5 degrees Celsius.

We already know the formula for morality that is small. M equals number off. Mordo Absolute. That means any divided by volume in leaders were given that the density of solution is one gram Parimal by mass. Total mass off 75 ml is already given and formula for Q solution. That means energy involved is equal to M c Delta T. We substitute the values Q equals 33.9 minus 25 degree senses multiplied with 4.18 multiplied with 75 grams. The energy involved will be 2.79 Killer Jones. It is known that Q solution is equal to negative off its reaction. So for reaction. So for reaction, the value will be negative. Two point 79 kilo Jones moving on to the next man. So we have 25 ml off. One more edge, too. S a four, which means 25 divided by close in Mullins is added to 50 ml off. One more load any which which can really presented by 15. Divided by comes in Moritz. According to the given reaction, One more off edge to a cell phone. The activated two moles off any witch so it can. So the only 25 by 1000 morals off any which is consumed in this reaction off any which is consumed. So the energy required what the reaction will be. Zero point 0 to 5 multiplied with Delta Edge. We know the Delta Edge offer. The action is a quintal que reaction divided by the number off moles off substance that is N. We substitute the values Delta X equals negative toe 0.79 kilo Jules divided by 0.0 25 Which gift? Negative. Triple 1.6. Hello, Jules. Therefore, the intel appear for reaction is negative 111.6 kilograms.


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