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HzO(g) co(g) = Hz(9) COz(g)Which of the following statements areJnd which are false.(alstAt equilibnumoycral compositionthe reaction mixturechanging slowly-H2O(g) t...

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HzO(g) co(g) = Hz(9) COz(g)Which of the following statements areJnd which are false.(alstAt equilibnumoycral compositionthe reaction mixturechanging slowly-H2O(g) two vessels are equivalent:co(g)vesselandH2(g) andcOz(g)aceganotherequz volume At equilibrium350*C _ the amountsco(g)Amounts of all reactants Jnd products corresponding Hrec equilibrium composition for this reaction Jre sealed period of time AC ould still be found only the CO molecules and not COz-vessel. The CO placed the vessel label

HzO(g) co(g) = Hz(9) COz(g) Which of the following statements are Jnd which are false. (alst At equilibnum oycral composition the reaction mixture changing slowly- H2O(g) two vessels are equivalent: co(g) vesseland H2(g) and cOz(g) aceg another equz volume At equilibrium 350*C _ the amounts co(g) Amounts of all reactants Jnd products corresponding Hrec equilibrium composition for this reaction Jre sealed period of time AC ould still be found only the CO molecules and not COz- vessel. The CO placed the vessel labelled with 14C. After an indefinite Ialse At equilibrum the rate the fonward reaction less than the rate the reverse teaction; Ialse Tne rate at which equilibrium established cannot be inferred from the magnitude the cquilibrium constant: equilibrum individual molecules be subject rapid chemicat trnstormations Nloced



Answers

Classify each of the following statements as true or false. a) Some equilibria depend on a steady supply of a reactant in order to maintain the equilibrium. b) Both forward and reverse reactions continue after equilibrium is reached. c) Every time reactant molecules collide there is a reaction. d) Potential energy during a collision is greater than potential energy before or after the collision. e) The properties of an activated complex are between those of the reactants and products. f) Activation energy is positive for both the forward and reverse reactions. g) Kinetic energy is changed to potential energy during a collision. h) An increase in temperature speeds the forward reaction but slows the reverse reaction. i) A catalyst changes the steps by which a reaction is completed. j) An increase in concentration of a substance on the right side of an equation speeds the reverse reaction rate. k) An increase in the concentration of a substance in an equilibrium increases the reaction rate in which the substance is a product. l) Reducing the volume of a gaseous equilibrium shifts the equilibrium in the direction of fewer gaseous molecules. m) Raising temperature results in a shift in the forward direction of an endothermic equilibrium. n) The value of an equilibrium constant depends on temperature. o) A large $K$ indicates that an equilibrium is favored in the reverse direction.

This is problem 98 from Chapter 16 and they give us a bunch of statements that we have determined, whether true or not so a saying we have a given temperature. All molecules help the same kinetic energy. Now, molecules up gas have different ranges of kinetic energy at a certain temperature rather than a given temperature. So a is false. He is half ing when we have half the pressure of a gases reaction, this doubles to right. So when the pressure increases, the collision decreases. And this causes a right of reaction to increase with the last number of collisions between the molecules, so he is also false. Okay, See, is that a higher activation energy gives a lower reaction rate. So when you have a higher activation energy, less molecules are not able to cross that energy barrier. So there actually is a decrease in the RIA reaction, so C is true, he's asking. A temperature rise of 10 degrees Celsius will double the rate of any reaction, so raising the temperature about 10 degrees Celsius for reaction. The number of molecules with the minimum energy increases so this will actually cause a rate of reaction to double. So he is true. He's asking when Monica was collide with greater energy to the activation. Energy distrust each change into product molecules. So when molecules collide, the energy and orientation work better when products are formed. So E is both. Okay, so efforts stating activation energy of reaction depends on temperature. So the activation energy definition is the minimum energy needed to start a reaction. And it depends on the catalyst. Which house speeds up the reaction by lowering activation energy. So it's not depending on temperature on temperature, so F is false. He's asking the rate of reaction increases when the reaction proceeds, so the concentration of reaction affects the rate. So if you have ah, decreasing concentration, this will cause a reaction rate to also decrease. So this will not increase. So G is folks. They just say, instead of activation interview this time, maybe it depends on a collision frequency. So this goes back to F when we were staying at the activation. Energy depends on the catalyst, which speeds up the reaction by lowering the activation energy So agents folks makes you have I and it stays. The callous increases the rate by increasing collision frequency. OK, so like we said before, the callous increases be for action by dropping the activation energy, this won't have to do anything collisions. So I is false, actually, have Jay Gay saying, except for Mick reaction there faster than endo. Thermic reactions. Eso thermic reactions release heat and endo. Thermic reactions absorb heat and absorbing the heat requires more energy to overcome these barriers, so exo thermic reactions don't need this additional energy. So Jay is true. Yes, we have. K K says that temperature has no effect on the frequency factor. So when there's an increase in collision collision frequency from an increase in temperature, this will cause an increase in collision between the molecules. So K is true rights. We have, l. I'll say, is the activation energy of reaction is little bite a cow? So, yes, as we mentioned before, this is true activation. Energy is the minimum energy. It's just our reaction. And this depends on the catalyst that Steve's of the reaction by lowering it so l is true. Thank you have any in state for most reactions, the change in age or in therapy is lowered by a catalyst, so callous speeds up the, um, activation energy by lowering it. So this isn't have anything within four p. So this is folks some reason it's gets in. So I'm gonna go, Oh, birth. You want to know And which true? Um oh, States, the initial reaction is the maximum. Great. So the initially definition is the maximum rate at the start of reaction. So oh, is true. Makes me happy. He states the bomb a bottom electoral reaction is twice as fast, and I, you know, military action. So for bottom, look, we have two molecules collide with each other, and this is better due to the energy and orientation between them. So yes, by Malika is 20 s past. So it's true, the last one we have this. Q. And he says the molecular parity of an element reaction is proportional to the military complexity of the reaction. So a definition of molecular parity is the number of reactions to form a product, So this isn't really have to do with anything with complexity. Sochi whistles

Yeah. In this problem we have to write overall reaction and red expression for the given mechanism. This is going to become in part a Given that two moles of a reactive with B to give the this is step age fast and reaches equilibrium. Mhm. Any step to the reacts with B to give E and F. This step it's slow. Any step three if give G. And this step is fast suppose rate constant four. These steps are Cuban. Okay. In verse care to and K three now to write overall reaction ID. This individual step and eliminate common spaces which occurs both sides of the reaction. Okay, on Irving. Well people go the kid. Mhm. Okay. two moles of a the activity B plastic plus B plus F. To give the E. F and G. Here we find dad, if is present in both sides of the reaction. So eliminate this species and these also present in both sides of the accent. So eliminate the we get two moles of a, reactivate too much of B to give E. And G, complicated. Therefore, over on reaction is two moles of a reactivate two moles of b. To give E and G. Now we have to write rate expression for this mechanism. We know that slowest step is the regular mining his step and in this mechanism step to its slow. So this step is regular mining step going to a little bit. Yeah. According to Yeah. Is it determining step? Oh yeah rate is proportional to the concentration of reactant present in this step. Therefore ready was to kato and took concentration of D. Into concentration of B. What we observed that D. Is not appeared in overall reaction. So we have to eliminate the from our death. It's prison. The first step is fast and reaches equilibrium. So values this is step to eliminate the we know that. Yeah. At equilibrium. What the system. Yeah. Yeah rate of forward reaction. Mhm. Is equal to rate of reverse reaction. Yeah. Mhm. So for step one we can right given into concentration of areas to the power to into concentration of B. Is equal to. Okay in verse into concentration of the So we get concentration of D. Is equal to given divide by the universe into areas to the power two Into Variously. Power one. Now substitute the value of D. In equation one we get oh red equals two. Care to into here, concentration of B is equal to given divide by K. Universe into concentration of areas to the power to in to be into B. Now replace care to given divided by K. Universe by experimental rate constant K. Into areas to the power to into the areas to the power to. This is our great expression for this mechanism. Don't give people hope. Mhm. Mhm. Mhm. She's coming. Yeah. Mhm. True. Yeah. Yeah. Now in part B. But given mechanism is a react with the to give see which is fast and reaches equilibrium. In second step C reacts with the to give F. This Jackson is also fast and reaches equilibrium And in 30 step f gives G. And this is slow. Consider this step one. This is a step too. And this reaction early step three suppose kevin and cain was a rate constant for this accent. Okay. Two and okay to universe is yet constant for this reaction. And K three's rate constant for this reaction we know that slowest step is the editor mining step. Therefore we can ride zip equals two. Yeah. Mhm. Yeah. K. three into concentration of F. Now, first we had to write overall the accent and overall reaction can be obtained by adding all these three steps. Have you seen club on adding? Mhm. All the evils three steps we get A plus B plus C. Plus D. Plus F. Gives see less F. Let's see here we find that C. N. F occurs both sides of the reaction. So eliminate these spaces. We get A plus B plus D. Give G. Therefore yeah. Overall reaction is A plus B plus D. Gift G. Yeah. Have you seen anything? Mhm. We have to ride. Mhm. Rate law for this mechanism. Don't take shit. Consider this as equation one here with the help of oral reaction we found that concentration of F. R. F molecule does not appear in overall the accent. So we have to eliminate this term if from rate X. Prison. This is from step one. Uh huh. We can write great off forward reaction. Yeah. Is equal to great off reverse reaction one. So we can write K. one concentration of a. into concentration of the is equal to okay in worth concentrates enough. See? Thank you. So we get concerned orthodoxy is equal to Cuban divided by K -1 concentration of A. Into concentration of B. Okay from step two article Librium rate of forward Jackson is equal to okay too. Into concentration of C. Into concentration of D. Is equal to K -2. Into concentration of if so concentration of N. Is equal to K. Two divided by k minus two. Into concentration of C. Into concentration of the Mhm. It's nothing Considimes us this education to consider this as equation three. Mhm. Yeah. Yeah. Yeah. Yeah no subsidy. Yeah we're gonna okay no substitute there. Value For concentration of f equals two K. two divided by K -2 Into C. in two. D. In equation one we get great equals two K. Three into. Okay to divide way k minus two, concentration of seeing too constant is enough D. Mhm. Important. We know that sees also does not appear in overall the accent. So we have to eliminate this. See from here mm We have created the value of C. Substitute this value kevin divided by k minus one concentration of a. Into concentration of B. Into concentration of D. Yeah we have such a truth value concentration of the from a question two we get experimental led constant care times concentration of A into concentration of B. Into concentration of day. Therefore required great expression equals to rate into concentration of Yeah into being to the with multiplication. Great constant care. Mhm. This is a great expression for part the Yeah what were those things? This is for part

11 so far this question, the reacted caution que has to be deter mined, which is two equals to react and upon product. So her QE was too page two, seal upon estudio. So the exponents are determined by the coefficients in direction. So substituting the values of the concentrations in the equilibrium constant, we'll get you equals to zero point eat in 20.25. About 0.5. So we'll get zero point to a 0.0.0 point five jay, Z goes to 0.4. So since Casey is equals to 0.5, so Q is great less than Casey. So therefore the forward reaction will take place. It indicates that there are more directing products, debt and then products. And the forward reaction will proceed to establish the equilibrium No. For these sub parts, we have to tell true faults. So first is false because the reaction mixture is not at a club in because he is smaller than Casey no. B. So it is also false because the forward reaction will take place. It indicates that there are more reluctant than products, and the forward reaction will proceed to establish equilibrium C is true because the forward reaction will take place. It indicates that there are more reactive than than products, and the forward reaction will proceed to establish the equilibrium while the D. Part is false, since Q is smaller than Casey, the forward reaction forward rate of the reaction is larger than the divorce rate. The forward reaction and the divorce rate are equally, are equal only at the equilibrium. That's the explanation, thanks


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