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Iron metal is dissolved in strong hydrochloric acid by the following reaction with a Kc value of 40.91 at 298 K. For the problems below, assume you begin wilh 5.0 g...

Question

Iron metal is dissolved in strong hydrochloric acid by the following reaction with a Kc value of 40.91 at 298 K. For the problems below, assume you begin wilh 5.0 g of iron metal and 100 mL of 0.5 M HCI and that the total volume of the reaction is unchanged with addition of metal:2Fe(s)6HCl(aq)2FeCl;(aq)3H,(g)Write the equilibrium constant expression.Hydrogen sulfide gas decomposes to hydrogen and disulfur gas with Kc value of 67 * 107 at 800 'C. If a 500 mL round-bottom flask initially con

Iron metal is dissolved in strong hydrochloric acid by the following reaction with a Kc value of 40.91 at 298 K. For the problems below, assume you begin wilh 5.0 g of iron metal and 100 mL of 0.5 M HCI and that the total volume of the reaction is unchanged with addition of metal: 2Fe(s) 6HCl(aq) 2FeCl;(aq) 3H,(g) Write the equilibrium constant expression. Hydrogen sulfide gas decomposes to hydrogen and disulfur gas with Kc value of 67 * 107 at 800 'C. If a 500 mL round-bottom flask initially contains 0.200 mol of HzS gas, what are the Hz and Sz equilibrium concentrations? 2 H,S(g) 2 Hz(g) Sz(g) Write the equilibrium constant expression: Write an ICE table for the reaction. Solve for the equilibrium concentrations_



Answers

Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas.
(a) Write the expression for the equilibrium constant $\left(K_{c}\right)$ for the reversible reaction
$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=98.7 \mathrm{kJ}$
(b) What will happen to the concentration of each reactant and product at equilibrium if more Fe is added?
(c) What will happen to the concentration of each reactant and product at equilibrium if $\mathrm{H}_{2} \mathrm{O}$ is removed?
(d) What will happen to the concentration of each reactant and product at equilibrium if $\mathrm{H}_{2}$ is added?
(e) What will happen to the concentration of each reactant and product at equilibrium if the pressure on the system is increased by reducing the volume of the reaction vessel?
(f) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?

Yeah. Okay, so this is a equilibrium question. So the equation chemical equation are given here, which is iron oxide act with hydrogen to produce the iron metal and water vapor. It also gives the entropy of this reaction. As we can see, the entropy is positive. So it's a and or systemic reaction direction will be accompanied by absorbing the heat. So the first question is right. The expression for the equitable constant. So we can work on that. The equation constant in this problem is equal to the concentration of the water vapor Power three over the concentration of the hydrogen gas power three. We don't have to involve iron or iron oxide because they are all solids. So for the solids we usually regard them as one in the equivalent constant expressions. So now we can move to the same question, What will happen to the concentration of each reactant and product equipment If more iron is added? So we can first because when we determine whether react the reaction move forward or backwards, we compare the reaction cogent with the queen constant. So again first right down the expression of the reaction cogent, which is actually saying as a queen constant. But this time the concentration is at any time any case not in the equipment. So you'll find the reaction cogent actually has not seen to do with iron because iron is solid. Which means by changing the mass of the island, we're not influenced change influence the direction Colton's. So the Q. C. Well remain constant and it will be still equals to the K. C. Their reaction equipment constant. In this case the requirement will not chance. So the concentration of the hydrogen and water will remain the same will be the same. It does not change. So now looking at the question, see what if what will happen to the concentration of each reactant and product a queen. But if water is removed, so them right on again, the cartoons of this reaction. So in the initial case we know the Q. C. Equal to the K. C. Right So it's in equipment state. But once it says we remove the water which means the concentration of the water were actually goes down. So instantly the newly uh uh reaction cogent well goes on. And this case it will be smaller than the K. C. Which means in this case through action would actually move right. Words move forwards. So in this case the hydrogen will be consumed. So the hydrogen concentration will be reduced and you are mainly remove the water. So definitely the water. The concentration of the water would be also removed. And now we can go to the question, what will happen to the concentration of each reactant and product equivalent brim? Ive hydrogen gas is added so we can treat it similar with the question. See we write down the reaction Collins expression again. Okay, and if we you see if we increase the month of the hydrogen, then the newly reaction cogent which actually goes down and initially it's equal to K. C. But after you remove the hydrogen actually would be smaller than the reaction um the queen constant and in this case it will be same through action will move forward, which means the water will increase and you are deliberately manly increase the concentration of hydrogen. So definitely hydrogen concentration of hydrogen will also increase. So now the question is what will happen to the concentration of each reactant and product at queen bee? Um If the volume of the reaction vessel is decreased. So this case you have considered the volume right? Yeah. So when you think about volume, you have to say the value change before and after the reaction you will find for the gas finish here, the reaction has three more molecules which is three more is hydrogen and also for the product that's also three more the um water. So which means for the reaction volume does not change, which means the value we have no influence on the equilibrium. So in this case the volume will not change the queue. He remains the same and the concentration of hydrogen and water does not change. And finally, yeah, is what will happen to the concentration of each reactant and product is the temperature of system is increased when you increase the temperature. So remember in the book, why says you can actually regard the heat as a reaction or product based on whether it's endoscopic or exhaust summit. So you see the delta H entropy is positive. So it's and atomic reaction, which means actually will absorb heat during the reaction. So we can add heat in the reaction site. In this case when you increase the temperature, which means to increase the concentration in the heat so definitely the reaction will move right. Words move forwards. In this case the reactant hydrogen would be consumed so the concentration will go down and the product award as water will be generated so the concentration will go up and this is the answer for the whole question.

Part, eh? Let's calculate the cake p using this formula here K c r T Delta End were given case. Uh, sorry. Calculate Casey given KP KP is 3.45 Casey is there unknown. Ours are ideal gas. Constance Temperature is 500. Calvin Delta End his morals of gas on the products mine. It smells of gas. The reactant switch should be two moles of gas in the products. One mole of gas and reactions to minus one is one. Solving for Casey gives us an equilibrium constant of 0.841 We'll use up for part B and part B. Let's for soul for the polarity of siege. Three. Three c c l in your show on. Initially, we were told that we have one more her five leaders, and that's going to be equal. The 0.200 muller set up a nice table. Ch 33 c c l gas in the equilibrium with ch 32 to gas and HCL gas. The initial amounts change equilibrium. The initial amount here is 0.200 00 minus x plus x plus x 0.200 minus x, x and x. Our equilibrium expression here is going to be defined. His products of a react in ce polarity the H three twos two h ugo over no u c c e o. We used the equilibrium constant that we calculated in part a 0.841 is equal to x x for the products 0.200 minus X for the reactant. So when we saw for X, we will get to roots. We will get 0.943 and negative 0.178 We're going to reject the negative roots, and we can solve for the equilibrium concentration of all four species. So if we do that, uh, we're going to define the concentration of see each 33 c c o. At equilibrium is going to be 0.20 Israel equilibrium, uh, the ice table to find that it's 0.200 minus x minus 0.943 and out with equal through a 0.106 Moeller the equilibrium concentration of ch three to see siege to we define is just X, so that would be 0.0 943 Moller and the validity of HCL ed equilibrium is also define us. Just acts So. 0.943 Moeller for Part C. We're dealing with partial pressures. So let's set up into the race table here and SPC H three three C c e o. Equilibrium with age 32 Si si tu in each CEO, and we're told that we have zero point for 0.6 atmospheres, respectively, minus X minus X plus. Acts and equilibrium lines finished a stable like so us to find our KP expression. Partial pressure of C three to C C, too. Partial pressure of each yell over the partial pressure of ch three. Three c c e o of substitute our values in From the beginning of the question, the KP value was given to us his 3.45 0.400 minus X 0.6 years or minus X and just x in the denominator solving for X will yield to roots will get Reid that is 4.39 and a second rate that is 0.546 We're gonna reject this route here to not have negative values and Let's go back to her. A stable. The partial pressure of C H 33 CCL at equilibrium we define is just X. This would be 0.546 atmospheres. The partial pressure, uh, C h three to see see each to his equal to we defined his 30.400 minus X, which will be equal to 0.345 atmospheres. And the partial pressure of this is equilibrium here, hcl it equilibrium was 0.600 minus X, which is equal to 0.545 atmospheres.

23 The reaction given is B r he is We'll call br Those powerful feel good went to help with the standard free energy change that ideology Nord which is no which is n for the liturgy, not off all the problems you don't have off all the products minus and their stuff, do you not my older The evidence It is half Delta Gino D f Well, we are too plus half their touch. Do not have feel too minus day rejecting that ideology. Note of off the RC two They are They're little spheres in your home. So after putting all the values using Appendix Day, we're going to that 0.54 killer jewels five months. So with a relation between equilibrium, that is K and standard free energy, we're going to calculate K. Uh, okay, then care is started. Stuff she not by rt when she is 2.54 Kill a jewel Farmers. My art is a 0.314 going to tend to the power minus three. Just remove the Calvin. His killer Joel farm. All Calvin on there is to 90 days. He killed them. So it was a 1.2 So, Allen case Okay, will be Ellen inwards. Ellen, Inverse off. One point. There are two. Which will we eat? The bar minus 1.2 or chill turns out the base 0.36 This is a value off carriages. The equilibrium constant. The initial concentration on we are sale is obtained by taking ratio of the moles and volume. The little would be RC else. One point it is little moles. Well, they are CIA. My 10.0 liters, which will be 0.100. Come. They live in constant expression for the reaction. Can Britain s care we are to, huh? Sale due? Uh huh. By older? Yes, sir. That is their serial. They are still sealed, so Yeah, so we need to find a change in the concentration eggs. So when it is of strategic kilogram concentration in this reaction okay, will be point receive. Have my to that's where to? To the bar house. My 0.1. Double zero minus six. Therefore, from here we can find as physical stool 0.41 name to convert most from the majority will take lie. We need to multiply the morality with this volume in letters. So 0.419 moles into 10 point zero letters will be 0.4 to 0. Well, the initial amount off reaction the initials amount. Well, for the yes Turned is point for one little little M in tow. 10 liters, which is equals two one. Well, the equilibrium, all the equilibrium, most off B R C l is one point. There is a minus 1.0.42 more. They are cycles 2.58 more off B. R. C s. So another collision concentration off the are too. His 0.42 moles. No, they are my two, which is 20.21 Mole? No, the arto. Now they live them constant off the alto, which is 0.42 moles off sale toe by two. Which is you don't point to one. No field. Therefore the amount off their cell tow is Montel be are safe. How's this point? Five Admin. The amount off they are to and point to one mall and the amount of sales. So this point to one more

11. So the first equation for the reaction is as follows, three FV plus four as two on equilibrium gives F. E. +304 plus four as two. And there Casey is four points is to find the concentration of H two. First, the calculator first calculate concentration of S. Two needs to be found. We have 24 g of S. Two and its molecular masses 18 0.2 g per mole respectively. So no for the moles 24 Grammys tool into one mall of as to oh a bond 18.0 to come a stool which is he goes to 1.3 mold of a stool. Once the mole are obtained, the concentrations need to be found using morality equation. The morality equation is more soldiers upon leader solution. So moral morality for immortality was two months old, salute is 1.3 mol and the latest solution is 10. We will get your point 13 more potato. So to find the concentrations at equilibrium the now we'll have to express the equilibrium constant for the reaction. So Casey was to H two four upon is too it is easily understandable. So the exponents that are derived by the coefficients in the reaction. Now substituting the values we'll get okay equals two four X. Well consider it for X. Upon 0.13 minus four decks pilot food substitute the value of case equals to 4.6. To get 4.6 equals to four X. I would fall upon 0.13 minus four X. To the power food. So taking fourth root of the equation four X phone 0.1 33 minus four X. It can be also written us to solve X. S follows 1.46 record 0.13 minus four X equals to four X. Which is it goes to 0.1 90 minus 5.84 Exit was 24 X. So 5.84 X plus four X equals two 0.190 after it moving further 9.8 for the exit was too 0.1 90. Where X equals two 0.4 90 upon 9.84 the X equals to zero point 393 So the concentration of H two erratically room is given by. It's Too equals to four X. So now four into X. The value of X. We have found 0.0 and 93 Real good approximately 0.77 M. Therefore the concentration of the H two at equilibrium is a little 0.77 Thanks.


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