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LS is positive forthe reaction Selected Answer: C3h8 (@) - 50 2 () _ 3 CO 2 (g) - 4 H 20 (E)Answers-2NO (3) 02 (3) ~ ZNO2 (3)C2H4 (g) - Hz (g) _ CzHs (g)3H2 (g) ~ 2...

Question

LS is positive forthe reaction Selected Answer: C3h8 (@) - 50 2 () _ 3 CO 2 (g) - 4 H 20 (E)Answers-2NO (3) 02 (3) ~ ZNO2 (3)C2H4 (g) - Hz (g) _ CzHs (g)3H2 (g) ~ 2 NH3 (3)2N2 Mg (s) - Cilz (g) _ MgC z (s)C3Hg (3) 5 02 () _ 3 COz (3) + 4 HzO (g)

LS is positive forthe reaction Selected Answer: C3h8 (@) - 50 2 () _ 3 CO 2 (g) - 4 H 20 (E) Answers- 2NO (3) 02 (3) ~ ZNO2 (3) C2H4 (g) - Hz (g) _ CzHs (g) 3H2 (g) ~ 2 NH3 (3) 2N2 Mg (s) - Cilz (g) _ MgC z (s) C3Hg (3) 5 02 () _ 3 COz (3) + 4 HzO (g)



Answers

Which of the following reactions is/are endothermic? a. $\mathrm{CO}_{2}(s) \rightarrow \mathrm{CO}_{2}(g)$ b. $\mathrm{NH}_{3}(g) \rightarrow \mathrm{NH}_{3}(l)$ c. $2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)$ d. $\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2} \mathrm{O}(s)$ e. $\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{Cl}(g)$

Problem 42 entrepreneur chemical kinetics and equilibrium. So in this ocean the formation of product is favored by decreasing the temperature are volume from which of the following reactions. So the concept and the question is the lady chatterley's principle. So according to the late chocolates principle, if a system at equilibrium is subjected to a change of any one of the factors such as concentration pressure or temperature, then the equilibrium is shifted in such a way as to nullify the effect of change. Okay, so in the first case we have given two months of sulfur trioxide equilibrium with the two months of sulfur dioxide plus oxygen gas minus it. So now, as you can see this negative saying of it in the product said so which means there is the absorption of the energy and when there is the absorption of the energy, this type of reaction is known as the and a thermic reaction. So for example, like if we say nitrogen plus oxygen, our equilibrium with the two more months of nitric oxide -8. So this shows that this is Sam and affirmative action. So now let us uh calculate the number of uh changing the number of cases most shelter and G. Which is the difference in the number of modes of product minus number of months of reacted. So these are the gassy smells. So I can get three minutes to that is equal to one so delta and she is the difference in the number of yes, a small so product and a number of cases small self reactant. So in this case the delta and develop greater than zero. So the reaction will product forward or proceed in forward direction only when decreasing the pressure or increasing the volume. So now let us discuss the next case. So uh again, this election is an anatomical action. The delta and the value that is zero. So in this case by increasing pressure or decreasing pressure have unchanged. No effect on the shifting of the equilibrium. So now in the next case we have four months of harmonia plus five months of oxygen. Our equilibrium with the four months of nitric oxide plus six months of water. So here Delta H. Is negative. So change in entropy value is negative. So which means this election isn't eggs atomic direction in which the it is released. And for eggs a thermic reaction, delta X. Value less than zero. Or we can say it's a negative. So delta and G value that is then minus nine. That is a cold often. So here delta and key value Correent greater than zero. So it means the reaction will go forward only when pressure is decrease or volume is increased. So then only let go. And the forward or weekend city product for water. Okay, so in the next case we have two moles of nitric oxide plus one mole of oxygen. Are equipped with the two moles of nitrogen dioxide. So here again delta H. Value is negative. That means the reaction is excited me as he is released and this delta H value negative for the extra thermic reaction. And for this reaction number of cases smalls that is equal to two minus three, which means minus one. So delta and G value here less than zero. So for this case if if we increase the pressure, yeah or we can say decrease the volume then the reaction will go forward. So increasing pressure or decreasing volume. So as in the question said that decrease in temperature or volume. So that meant for this case before this match with the so this much with the question ask. So that is the decrease in volume. So hence I can write formation of products favor by decreasing the temperature or volume or volume as because uh again this is the eggs atomic reaction for delta H. Value negative. So for this reaction in case of eggs atomic reaction if temperature is decreased, then again this reaction is the product. So therefore from the given option option D. Is the correct. So I can write option the which is two months of nitric oxide. That's one mall of oxygen gas. Keep them with the two miles off nitrogen dioxide, then delta X. Value is negative. Delta H. Value is negative again for this case. So I can write, the heat is at least. So I can also write in this way this reaction, so hence you can see formation of product is favored by decreasing the temperature, decreasing the temperature or volume.

So energy plays a key role in chemical processes. And according to the modern view of chemical reactions, bonds between atoms and the reactions need to be broken and the atoms or pieces of the molecules can be reassembled into products to form new bonds. So here we have two moles of 02 That reacts with eight moles of N. O. And this gives us four moles of N 203. In the next example, we are changing the striking metric values so we have not point not 23 moles of N. O. That can react with not not 575 moles about two To give us not .115 moles of N 203. And so we can calculate the number of moles of N reacted. So the number of moles reacted is the weight divided by the Molar mass to have 4.73g, divided by 14.01 at 16 zero g per mole, Because this is nitrogen and this is oxygen. The molecular weight from the periodic table. We got not .236 moles, and so finally, what we have is No .236 moles of n. o reacts with not not 59 moles of two together, is not .118 moles of N 203.

Solution for the evil problem where we have to calculate Delta Jeannot for the glove innovations. A 25 degree Celsius so calculation for part a Delta Jeannot is equal to two Delta G. Note. If not into nitrogen oxide minus delta. Achy not into nitrogen molecules minus delta. Achy not into or two molecule captain. By putting values we have toe into 86.7, minus zero minus zero. Kate, you will Permal. Is it well too? 173.4 killed with Joel Permal? Is he calculated value in next four, but be Delta Gee. Note is equal to Delta Jeannot into H toe in this c state minus Delta if not in tow. H toe in legal state by putting values behave. One in two minus 2 to 8.6, minus one into minus 237.2. Killer. Joel Permal It's equal toe 8.6. Killer. Joel Pirmin Venture For part C. Delta Jeannot is equal to four Delta, if not into C 02 plus two into Delta, if not into H 20 minus two into Delta. Achy not into C two h two. My news five into Delta G not into or two By putting values. We have four into minus 394.4 plus two into minus 237.2 minus to multiply 209 minus two minus five. Multiply zero is equal to minus 2470 Kelo! Joel per more is the calculated offensive. Uh huh, Yeah, yeah, yeah!

So let's take a look at when K P and K C can be the same value. So let's go back to our equation that relates the two K P is equal to K C, times R T to the delta and G. And so the key here is the delta and the not changing number of moles of gas in the reaction. Because when uh, delta and G is zero, right, then something to the zero power is one. And that winds up making this whole term, if this delta and a zero, that makes this whole term one. And so one times K C is the same as K P. So when delta and is zero, that is when we're going to have K P equal to K C. So we want to see the change in number of moles of gas to be zero. So the number of moles of the product side, minus the number of moles of the reaction side has to equal zero. So if we were to look at several chemical equations and these are all gases, we would see that in this first example, we have two moles of gas on the right, two moles of gas on the left, so the change in number of moles of gas is zero. So that means a K C and K P are going to be equal. If we were to look at another chemical reaction, Went to add some cl 2 to that side. Well, in this case we have one mole of gas on the right, two moles of gas on the left, 1 -2 is -1. So if we were to plug negative one into that exponent K C and K P or not equal. And finally, if we were to take a look at a third reaction again, everything is gashes to N 02 We would see we have two moles of gas minus one, and so that is not zero. So Casey is not going to equal k p, k c equals k P only when delta and a zero because anything to the zero power is one.


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