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Nz(g) + 3H2 (g) = 2N Hz(9) Kcq = 6.7 x 10-6 The reaction above is run; the concentrations of Nz, Hz and NHz measured and recorded in the table below:Nz] [Hz [NHz]9....

Question

Nz(g) + 3H2 (g) = 2N Hz(9) Kcq = 6.7 x 10-6 The reaction above is run; the concentrations of Nz, Hz and NHz measured and recorded in the table below:Nz] [Hz [NHz]9.25 x 10-2 M 8.27 x 10-4 M 2.23 x 10-5 MBased on the measured concentrations will the reaction proceed in the forward or reverse direction to reach equilibrium?the reaction is at equilibriumproceed in the reverse direction; towards reactants proceed in the forward direction; towards products

Nz(g) + 3H2 (g) = 2N Hz(9) Kcq = 6.7 x 10-6 The reaction above is run; the concentrations of Nz, Hz and NHz measured and recorded in the table below: Nz] [Hz [NHz] 9.25 x 10-2 M 8.27 x 10-4 M 2.23 x 10-5 M Based on the measured concentrations will the reaction proceed in the forward or reverse direction to reach equilibrium? the reaction is at equilibrium proceed in the reverse direction; towards reactants proceed in the forward direction; towards products



Answers

Consider the reaction: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ Complete the table. Assume that all concentrations are equilibrium concentrations in moles per liter, M. $$\begin{array}{ccccc} \hline T(\mathrm{K}) & {\left[\mathrm{N}_{2}\right]} & {\left[\mathrm{H}_{2}\right]} & {\left[\mathrm{NH}_{3}\right]} & K_{\mathrm{eq}} \\ \hline 500 & 0.115 & 0.105 & 0.439 & \\ 575 & 0.110 & & 0.128 & 9.6 \\ 775 & 0.120 & 0.140 & & 0.0584 \\ \hline \end{array}$$

In this question, we have been given one of the main reactions in the harbor process where nitrogen reacts with hydrogen form more now. So what we have here, we first of all, I have to write the expression for casing and we know that Casey is equal to the concentration of the product divided by the concentration of the reactor. And we always have to make sure that we raise the respective concentrations. They asked the geometric coefficient. So in this case for this reaction, our casing Casey is going to be equal to the concentration of ammonia divided by the concentration of hydrogen and the concentration of metric. So we have to raise this to the power of to and patrick into the power this is hydrogen to the power of right. So in different cases in that chair phone we have one of these parameters that is missing. That is for the first one we have Casey that is missing. The second one we have H two that is missing. And for the final one we have concentration of ammonia that is missing. So all we have to do is to just make what is missing. The unknown parameter subject of the formula in this way in this equation of casing. So for the first part we have Casey we're supposed to complete casing and K. C. Is equal to the concentration of ammonia. This is the expression so our Casey is going to be equal terms 0.439 and we have to square this divided by the concentration of nitrogen and hydrogen which is the point 115 time zero point 105. And this is the way to the power of trend. So our Casey is going to be equal to 1.45 Times 10 to the power right? And then moving on to the next one where we want to calculate the concentration of hydration. So if we are to make hydrogen subject of formula here, we're going to have something like let's just isolate this. We're doing this training. So what we're going to have here is K. C. Multiplied by concentration of nitrogen and the concentration of hydrogen Best of three. And this is required to the concentration of And it's very raised to the power of two. So H two To the power of three is equal to an H three. The power to divided by K. C. Multiplied by concentration of matron. So at the end of the day, the concentration of hydrogen, this is going to be equal to the cube truth, The cube root of the concentration of I'm one year quit. Divided by Casey multiplied by the concentration of nitrogen. So this is going to be equal to the cube root of 0.128 Squid, Divided by 9.6 Turns opened. 1 1. So the concentration of hydrogen concentration of hydrogen in this case is going to be equal to zero point 249 in malls. Political own mall spain cubic dissonant. Now we do the same for net ammonia. Well, the concentration of ammonia is going to be equal to the square root of Casey, Multiplied by the concentration of N two and the concentration of attrition based on the power play. So the concentration of ammonia, this is going to be equal square road. This is the square at all. 01, Time, times 0.14 to the power. So the concentration of ammonia is the quality. Four point 3 9 terms turned to the correlated to today. And this is also in malls per litre, almost bare cupid. Yes, to me. So these are the missing values in the table where we have Casey concentration of hydrogen. Finally, the concentration of Albania.

Hello women. So in this question we have to complete the table, kevin table. So this is the chemical equation and now we'll have to express it for the equilibrium question. So we'll get case equals two NH three two upon and two H 23 So now equilibrium concentrations at 500 kelvin Calvin is foreign to. It's zero point when 115 mm. Whereas for H. Two it's about zero point 105 him. Whereas for NH three it's about 0.439 So upon substituting the data and they're following constant expressions will get K. C equals two zero point 439 square upon 0.115 in 20 point 105 cube. So we'll have to solve this. And after solving this, I've already solved. So after solving this we are getting when 0.45 into then issue the power three. So now at 5 75 75 Calvin Casey will be 9.6. That's good. So the end to will be zero point 11:10 a.m. Whereas NH three will be now 0.128 So now upon substituting the data in the equilibrium constant expressions. So we'll get 9.6 equals to zero point when 28 upon 0.110 into 9.6. As to cube is 9.6. Sorry? Sorry sorry sorry sorry. Sorry. It is H two killed. So we'll take esta que to the left hand side and 9.6 to the the denominator of the right hand side. So we'll get s two 31 point 0.128 square upon 0.11 zero in 29.6. After solving this will get around one point 0.1 If I five M. So H two zero point 249 All right. And now at 7 17 five K. This was for 5 75 75 years. No, 7 75 K. And you will be with 0.1 to zero. As to will be zero point 14 PM case. He will be is given a zero point 584 Now will substitute the data and therefore equilibrium constant expression. So we will get, you know, point deal 584 equals two. NH 32 power to 010 So 0.12 went to zero 0.140 cube. So now, after solving this, will get the value for NH three as 0.43 nine AM. And hence it is your answer. Well, I thought it should be No. Clearly Industry two year old. And thank you for watching

There are a serious of reactions and were given the concentrations for the reactions and products and we need to find the equilibrium constant. So to do this, K e que, in general equals the concentration of the products over the concentration of the reactions. So let's go ahead and start with the 1st 1 So OK, ik you for this one is equal to the concentration of C over construction of a times concentration of B and the coefficients are all one. So it's to the power of one which is just itself. And then we can just plug in the values because this concentration to this is moles per liter. So it's in the right units. So then we have 4.0 2.0 times 3.0 and this gives us, um 0.67 Then we could do the same thing with the rest. So okay, he que is equal to the concentration of g times the construction of F. But we need to account for the coefficients. So we put it to the power and then the reactions are e the part two times D, and then we can plug in our values. Um, so you get 1.8 times 1.2 cubed, 9.5 times two squared, and then we get 0.52 and then for the next one. So this is the same thing. We're given actual reaction. So an important thing to taken account is to make sure that they're all gases is only gases can go into the equilibrium constant. So then it's in each three squared, cubed times, the other reaction nitrogen, and then every plugging these values. And then that equals 311.3. We want to keep it to, say, figs. So we can also say that's equal to 3.1 times 10 to the second. And these are your final answers.

In this given chemical reaction, three molecules of a reacts with one molecule, er B to produce two molecule of C and one molecule of D. One thing we cannot hear that is in gaseous state. B is in pure solid estate, while C and D are inadequate state. Now, after the, after this reaction has achieved equilibrium, their concentrations in terms of mood or more clarity is given in the problem. So is 2.48 mol B is 2.41 mole C has the concentration 1.13 Mueller, and D has the concentration 2.27 Golden. So these are the datas are the equilibrium values Now in the problem, it is also mentioned that volume of the flask in which the a gas has filled. The volume is 1.87 liter. And we have to find equilibrium constant for this lee Jackson. So first of all, we will we will convert all the concentrations in a single unit, since C. And they are already in Mueller units. Therefore we will convert the concentration of a also in Mueller unit. So how Mueller unity is achieved for that. We divide the moral values with the uh most by volume. So here for a the mole is 2.48 moles divided by volume is 1.87 later. So this will be the modularity of I guess now the question is how we will write expression for Casey or equilibrium constants. We know that while we write the expression for equilibrium constants, we neglect the pure solids and pure liquids here. Pure solid is present and that is at um be therefore we will neglect its concentration since its concentration will not change with time, it will be always unity. Now, first we enumerated, we write the product concentration 1st. So what is the water? The products that are C&D. Now the power of these concentrations will be their corresponding coefficient. Now, coefficient of CS two, Therefore, concentration of C will be squired and coefficient of D is one. Therefore there will be one in the power of fish, in the power of the concentration of D. So this will be divided by the concentration of reactant and there is only one reactant since we are neglecting the concentration of be it, since it's a pure solid. Therefore, concentration of A. And its coefficient is three. Therefore it will go into the power of A. So this is equal to now we can put the data's of their concentration values at the equilibrium. So C. Is one point 13 Mueller police square and two D. Has the concentration 2.27 Mueller Divided by concentration of a. That is 2.48 divided by 1.87 whole cube. So from here we get the value of Casey that is equals two, one point 242 So the correct option is option E. That is one point 24 so this will be the answer for this problem.


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