Find the Ph of a 1.60102 M Acetic Acid Solution.
20) Which of the following A) ASuniverse statements = always TRUE? Entropy greater IS not than temperature = zero for _ nonspontaneous process. 20) C) Entropy is an extensive - dependent: D) Exothermic property_ E) None ofthe pbocesses - decrease " the above are entropy of the true. 'surroundings. 21) How many grams of KBr A) 5.65 are required to make 650.mL ofa 0.115 M KBr solution? B) 8.90 & 748. D) 0.628 22) For the first-order reaction, 2 NzO(g) Nz(g) 02(g), what is the conce
20) Which of the following A) ASuniverse statements = always TRUE? Entropy greater IS not than temperature = zero for _ nonspontaneous process. 20) C) Entropy is an extensive - dependent: D) Exothermic property_ E) None ofthe pbocesses - decrease " the above are entropy of the true. 'surroundings. 21) How many grams of KBr A) 5.65 are required to make 650.mL ofa 0.115 M KBr solution? B) 8.90 & 748. D) 0.628 22) For the first-order reaction, 2 NzO(g) Nz(g) 02(g), what is the concentration of N2o after 3 half-lives if 0.15 mol of N2O is initially placed into a 1.00-L reaction vessel? A) 1.9 * 10-2 M B) 3.8 * 10-2 M C) 7.5 * 10-2 M D) 94*10-3 M 23) A reaction is found to have an activation energy of 38.0 kJlmol. If the rate constant for this reaction is 1.60 102 M-Is-1at 249 K what is the rate constant at 436 K? A) 3.80 = 104 M-Is-1 B) 2.38 * 105 _ M-Is-1 C) 4.20 105 M-ls-1 D) 1.26 103 M-Is-1 E) 7.94 104 M-Is-1
Indicate whether each statement below is true or false. If a statcment is falsc, rewrite it to produce a closcly rclated statement that is true. (a) For a given reaction, the magnitude of the equilibrium constant is indcpcndent of tcmperature. (b) If there is an increase in entropy and a decrease in enthalpy when reactants in their standard statcs arc converted to products in their standard states, the equilibrium constant for the reaction will be negative. (c) The cquilibrium constant for the rcversc of a reaction is the reciprocal of the equilibrium constant for the reaction itself. (d) For the reaction $$ \mathrm{H}_{2} \mathrm{O}_{2}(\ell) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\ell)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) $$ the equilibrium constant is one-half the magnitude of the equilibrium constant for the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(\ell) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{O}_{2}(\mathrm{~g}) $$
Let's take a look at several reactions uh and take a look at several statements and see whether they might be true or false and if they're false, what we can do to correct them. So the first, let's take a look at uh part A or hydrogen and nitrogen become ammonia. And um, you know, do we just have to depend in this chemical reaction on the entropy of the system? And no, we don't. So remember if we want to know whether reaction is going to proceed or not proceed. We want to know about the Gibbs free energy, which is going to be about entropy, but it's also going to be about entropy. So we don't just have to look at the entropy. Entropy is also important. If we take a look at the second example where we have sodium a chlorine to make sodium fluoride. Um, you know, is this a spontaneous process? Well, the way to be able to figure that out is if we knew it gives free energy. And so if we took a look at the gibbs, free energy of reaction, right? Which is the sum of the number of moles of the gives energy of the products minus that of the reactant. Since what we know is that elements in their natural state, like N. A. And C. L. Two. They have gibbs energies of formation of zero. So it's only going to be related to the two uh moles of the N. A. C. L. Which has a gibbs. Free energy of negative 384 killer jewels per mole. And then we got 0-plus zero for the react inside. And so we have a gibbs free energy of negative 768 killer jewels, negative is spontaneous. So that makes this one true. And if something is spontaneous, does that make it reversible? Yes, it can be reversible. This is true and it's something spontaneous. Does that mean that work gets done? No. Uh If we need work to get done, that has to be a non spontaneous process. And so if that's the case this is false. And our spontaneous processes always eggs a thermic. Well, no, not always. But they could be so remember again that we have our gibbs free energy which has to be negative if it's going to be spontaneous. And so we could have an X. A. Thermic process where DELTA H. Is negative, but we could have a positive delta H. And we just need a very positive DELTA S. To get this term to be negative enough to make the delta G negative. So if we have something that creates a lot of entropy that would also be spontaneous. And that makes this last one false.
For discussion. We have to tell whether the provided statement is true or false. So auction, eh? In this case, the statement is false because the manufacturing off ammonia does not entirely depends on the value off Delta H. So the lead off forward reaction option B is true. Option C is also true. Auction D. We have falls because non spontaneous process require some work to be done. Lastly, via Auction E, which is also false. Because the spontaneous process are mostly eggs atomic but not all off the books. It increases the disorder off a system.
Of the reaction for the dissociation of Die Atomic A into two singular a molecules. And we're going to discuss some of the entropy and the P and energy properties of this reaction. So our first step here is to look at the signs our standard adults H and R standard. So the standard Delta US is a little bit easier to find. You can look right at this reaction equation and look at the number of moles on either side. Remember that if there's more moles of gas, the entropy of that reaction mixture will be higher. So let's look here. We have one more gas reacting on the reacting side, and then we have two moles of gas on the product side. Therefore, we have more Mel's on the product side, and the entropy of this forward direction reaction is going to be positive. To find Delta H. We want to look at these two reaction mixtures. Here we have initial state one with eight Die Atomic A's and two singular A's, and we have reaction. We have this reaction make sure equilibrium that has six Die Atomic A's and six singular A molecules so visually here you can see that majority of the die atomic A molecules are still in the reaction at equilibrium. This must mean that the reaction is endo thermic and requires an input of energy in order to fully dissociate die atomic A into singular A. And because this reaction is an author, Mick, that means our Delta H is greater than zero. It is positive. So for our next part here, we're going thio distinguish between what Delta s means versus Delta s standard. So our delta s standard so we can conceptually understand the difference between these two just from the idea of standards alone. So our Delta s standard is the amount of entropy in the entire reaction when associating a 21 mold A to completely converting it into two moles off singular A. So that is a standard for this reaction. Whereas this Delta H with the note or this Excuse me, this dealt, asked with the no standard is the entropy for the reaction at its current progress. So this number can change. This number cannot change. So for our third part of the reaction or third part of our question here we are talking about the sign of our Delta G standard. So when you think about Delta G standard, we can use our equation that Delta G standard is equal to Delta H Standard minus the temperature times the standard entropy of the reaction. So we ran into a little bit of a problem here. We have adult H. That's positive. We have adults, us that's positive. And that makes it almost impossible for us to determine the sign of our Delta G standard here so we can call up a few scenarios. So at points when our temperature is very high, that will cause our negative T Delta s term to increase. This in turn, will cause our Delta G standard to be negative at lower temperatures are negative. T Delta s term here will decrease. Therefore, our delta G standard will increase will be greater than zero. So now that we know are signs for Delta G, we can go ahead and look into our K p. So our KP is our equilibrium constant here in our reaction. So when we have our equation of Delta G standard is equal to the negative. Rt l n of Cape er Gibbs Free energy equation as we increase our temperature from up here. Remember, as we increase our temperature, our Delta G standard will become less than zero. It will become a negative number. Therefore, this entire term here, we'll need to be larger. So we can say that as temperature that was a Delta, uh, temperature increases R K p will also increase. So then, for our final part here, we're thinking about Delta G not standard. Just normal. Delta G. So far a delta Gee, at equilibrium when we're at equilibrium are Delta G is going to be equal to zero because there is no change in free energy at equilibrium. There's no change. This Delta G is going to be equal to zero, even though our adult that you standard or at different points of the reaction this Delta G value can vary when we're at equilibrium. Furthest association reaction are Delta G will be equal to zero
Question. Number 80 is also a very good review of the entire chapter. You were given several statements and asked to oh, identify them as being true or false. The first statement is some equilibrium depend on a steady supply of reactant in order to maintain the equilibrium. Now this is kind of a trick question, because we do need it a steady supply of reactant. But coming from the products reforming reactant, I believe this question is suggesting that we're supplying a steady, steady supply of reactant from the outside. We're adding in reactant from the outside. This this would be incorrect because that would be an open system and an equilibrium Onley occurs in a closed system. So this statement is false. Part B says that both of forward and reverse reactions continue after equilibrium is reached, and that's what defines the dynamic equilibrium, which is what occurs. So this is true, Part C says. Every time reactant molecules collide, there is a reaction that occurs. This is false because there are several things, or to at least two other things that are required for a reaction to occur after a collision occurs. One is there needs to be sufficient energy and to their needs to be proper orientation. So because of these two additional factors, not all collisions result in a reaction. De says potential energy during a collision is greater than potential energy before after the collision. This is true because if we draw the energy diagram, we will see that the greatest potential energy occurs at the transition state where the activated complexes, which is what occurs after our during the collision. So this is true. The potential energy during the collision is the greatest part e the properties of inactivated complex air between those of reactant and products. You might assume that this is true. It must be because we've got reactant and we've got reactant coming together to form product, and we don't have reactant anymore. We haven't quite made product, so it's likely that it has properties between the two. But we really don't know because we can't isolate the activated complex and measure its properties. So we'll have to say faults on this one. For F activation, energy is positive for the forward and reverse reactions. This is true activation. Energy is always positive. Fergie Kinetic energy has changed to potential energy during a collision. Yes, this is true as the kinetic energy of the colliding molecules. All right, the kinetic energy of the colliding molecules provides the potential energy to create the activated complex. So this is true. H an increase in temperature speeds the forward reaction but slows the reverse reaction. No one increase in temperature will increase the rates of both the forward and the reverse reaction because more collisions air occurring in the forward reaction and reverse reaction with sufficient activation. Energy I a catalyst changes the steps by which your reaction is completed. Yes, this is true. A catalyst provides an alternate route that has a lower activation energy. And it's because of this alternate route that the reaction can occur more quickly in the presence of a catalyst. Jay is an increase in concentration of a substance on the right side of an equation speeds up the reverse reaction. No, an increase in reactant is going to increase the rate of the forward reaction, not the reverse reaction. So I think I read that wrong and increase in the concentration of a substance on the right side, which is the product side, will increase the rate of the reverse reaction. Yeah. So this is true. An increase in the reactant concentration increases the rate of the reverse reaction. True. Okay. And increase in the concentration of a substance in an equilibrium increases the reaction rate in which the substance is a product. That's a mouthful that I don't It's not very clear, in my opinion. I don't know if you are able to interpret it better than me, so I'm going to restate it. An increase in the concentration of product increases the rate of the forward reaction. No, that would be false because increasing the rate are increasing the concentration of product would increase the rate of the reverse reaction, which is what I was trying to answer in the previous statement. I But anyway, it's okay. If you increase the amount of product, you're going to increase the rate of the forward reaction, not the reverse reaction. L Reducing the volume of, uh, in equilibrium system will shift it towards the side with fewer moles of gas, fewer molecules of gas. This is true. We have less volume, so we can't accommodate us many molecules of gas. So, um, we need to shift to the side with fewer moles or fewer molecules of gas. I am raising temperature results in a shift in the forward direction for an endo thermic reaction. Yeah, raising temperature for an Endo thermic reaction is going to increase the K value and in the process shifted in the forward direction. That's true, and the value of an equilibrium constant depends on temperature. Absolutely. We just talked about that in reference to this end, a thermic reaction. Increasing temperature for an Endo thermic reaction increases the K value and, oh ah, large K indicates that in equilibrium is favored in the reverse direction. Nope, that would be faults. A large K indicates that a direction is favored in the forward direction, not the reverse direction.
Find the Ph of a 1.60102 M Acetic Acid Solution.
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