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Suppose 250. mL flask is filled with 0.60 mol of Iz and 0.40 mol of HI. The following reaction becomes possible: Hz(g) +Iz(g) 2H(g)The equilibrium constant K for t...

Question

Suppose 250. mL flask is filled with 0.60 mol of Iz and 0.40 mol of HI. The following reaction becomes possible: Hz(g) +Iz(g) 2H(g)The equilibrium constant K for thls rcaction 4.07 at thc tcmpcrature the flask:Calculate the equillbrium molarity of 12- Round Your answcr t0 two decimal places:

Suppose 250. mL flask is filled with 0.60 mol of Iz and 0.40 mol of HI. The following reaction becomes possible: Hz(g) +Iz(g) 2H(g) The equilibrium constant K for thls rcaction 4.07 at thc tcmpcrature the flask: Calculate the equillbrium molarity of 12- Round Your answcr t0 two decimal places:



Answers

Assume you place $0.010 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{4}(\mathrm{~g})$ in a scalcd 2.0-L flask at $50 .{ }^{\circ} \mathrm{C}$. After the system reaches equilibrium, $\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.00090 \mathrm{M}$. Calculate the valuc of $K_{w}$ for this reaction. $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) $$

So here we are asked to determine the equilibrium constant with respect concentration. So even for gases we're going to use concentrations. So we have and die nitrogen tetroxide gas which is an equilibrium with two moles of nitrogen dioxide and were given information that initially there is a concentration of Die nitrogen tetroxide of there's essentially 0.01 moles of it and the two liter container. So we have to convert everything in terms of concentration. So 0.010, divided by two, would just be 0.005, The 0.0050 molars. And it's important to remember that we have a volume of two leaders And equilibrium were given information that 0.00090 Mueller's remain of the substance. So a certain amount must have reacted in this process And the amount that must have reacted with the 0.0041 Mueller's. So since a certain amount of the react and has reacted, that must contribute to increase the concentration of product initially we have no nitrogen dioxides or initial concentration is zero. However, two moles of nitrogen dioxide are formed per mole of Uh die nitrogen tetroxide consumed. So this would be two times 0.0041 And we result in 0.00 82 as our final concentration. So these are the concentrations at equilibrium. Our equilibrium constant with respect to the concentration would be this adjusting for strike a metric coefficients. Yeah. And now we can use this as a result to find our equilibrium constant for this reaction. Yeah. And we find as a result for this specific reaction, this reaction has an equilibrium constant of 0.075. And this is our final answer.

But here we have a specific reaction or die nitrogen tetroxide associates into two nitrogen dioxide. And were given that the equilibrium constant for this reaction is 5.8 times 10 to the -3. Were given the initial amount of dye nitrogen tetroxide, Which uh 15.6 g. And we essentially can first calculate the molar mass of Dina trajan tetroxide, which would be 28 plus 64. So 15.6 divided by that would give the mountain roles. And since we have a five liter flask, we have to divide by that to find the concentration Which we find is essentially 0.034 molars. Mhm. And we know that in the reaction there's going to be in a decrease in the amount of dye nitrogen texture oxide, increasing the amount of essentially nitrogen dioxide. And we have the following expression for our equilibrium constant. So we can temporarily make an assumption that X essentially much less than 0.034. And see if this assumption is justified. So as a result, we can solve and divide and take the square root. As a result, we see that we find X is about 0.0070 molars. And checking our assumption, Do we see that this assumption essentially not justified since this is too large relative. So we essentially have to apply the quadratic formula. As a result. We essentially find that using the quadratic formula X is essentially about 0.0063 molars. So this means that equilibrium, there's going for the concentration of nitrogen dioxide in terms of a concentration of nitrogen dioxide would be two times this amount And this would be equivalent to essentially a balance 0.013 molars. In terms of moles, since there's a century five liter, a five liter flask, we have to multiply by five and we can find that the amount in the century moles Is essentially equivalent to 0.063 moles And the amount of dye nitrogen tetroxide that's associated is equivalent to 0.0063 divided by 0.034, multiplied by 100. And we can see that the amount of associated is essentially about 18.5%. And this gives our final answers.

In order to solve for, uh the way the equilibrium shifts will use the equilibrium h 20 plus. He'll too. Oh, unequal room to a juicy L E equilibrium constants Expression is given, and we see that that's equal 2.900 Calculate Hugh, which is the quotient, or trial Kiki Value, which is based on the initial concentrations of each of the reactions from products involved in the equilibrium. For a we need to solve the initial values initial values are sold for by dividing each of them all amounts by the one leader container April CEO Initial eyes one Moeller seals who Oh uh, like Lawry a die chlorine oxide initial is 0.10 Moeller and the water vapor is 0.10 Moeller when we substitute these values into the que with three quotient which to find up above on substituting these values in and solving gives us a question Top 100. We take the question to compare this the actual equilibrium constant. When we go ahead and do that, we see that the quotient 100 is much greater than the equilibrium constant of 1000.900 Therefore, since Q is larger than K, the equilibrium shifts left for Part B. We have a different set of conditions initially, so we need to solve for the initial values, like so dividing each of them all amounts by two leaders. Because this is a two liter container. This time we take all of those values and we substitute those into the expression for the quotient, and that gives us a Q value of 0.9 zero. Compare this to the equilibrium constant. We see that this is exactly equal. Q is exactly equal to K, so in this case, if they're exactly the same, the system is at equilibrium. For part C, we again need to solve the initial values here. For the three. We take the three initial values you can see, given their dividing by 53 leader container. This time, substitute into the quotient expression, which gives us a question to a truck. He Q value of 110 Compare this to the equilibrium constant in 110 key was much greater than the 1100.9 Therefore, since Q is greater than K, the equilibrium shifts left

Setup. Equal the brim question here too. Oh, no! CEO equilibrium were to end old gas field Thio Theo The table here zero 2.1 point zero would be the initials. The only possible way here would be the shift to the left. The more Isha 2 to 1 two, x 2.0 minus two x 1.0 minus X and the equilibrium constant here would be, you know, squared seal to over an O. C. L squared. We're given the value for the equilibrium constant here, which is 1.6 times 10 to the negative five. You put zero minus two x points reminded Sex two X. Solving this equation for ex would give me a value here of 0.975 Take that back up to the ace table. And when we do so, that would give us equilibrium values for uh, N O C E O is two times 20.975 So this would yield 1.95 Moeller. The mill arat e of n o at equilibrium would be to minus two X, which would yield 20.50 Moeller and one minus X Would yield 10.25 Moeller for the equilibrium concentration of C l, too. So those would be the three days and equilibrium that we were to solve for.


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