Calculate the equilibrium concentrations at 25 °C for the reactio...

Question

Calculate the equilibrium concentrations at 25 °C for the reaction in Problem 15.84 if the initial concentrations are 
3N2O44 = 0.0200 M and 3NO24 = 0.0300 M.

Accepted Answer

hi everyone for this problem. It reads the following reaction has an equilibrium constant equal to 12.82 at 375 kelvin calculate the concentration of each species. Once equilibrium is established if the initial concentration of sulfur dioxide is equal to 3750.2 molar chlorine is equal 2.15 molar and sulfuric chloride is equal to 0. molar. So the question that we want to answer here is the concentration of each species once equilibrium is established. Okay, So what we're going to want to do is we're going to determine which direction the reaction will proceed by comparing Q two K. Okay, So in the problem we're given what the K value is, which is our equilibrium constant. So we need to compare our equilibrium constant to our reaction quotient Q. And if Q is equal to K, then the reaction is at equilibrium. If Q is greater than K, then the reaction shifts in the reverse direction to reach equilibrium, which means the reaction shifts left and when Q is less than K, the reaction is going to shift in the four direction to reach equilibrium, which means the reaction is going to shift right? Okay, so the expression is going to be the same that the equilibrium constant expression and reaction quotient expression are the same. So, if we want to calculate the reaction quotient, it's going to equal the concentration of products over the concentration of react ints. Okay. And we need to take a look at our equation to do this and we have all gasses species. So this is good. So our reaction quotient then is going to be the concentration of product over the concentration of react mints. Okay. And we're given these values these values were given in the problem here. So all we need to do is plug these values in so that we can solve for our Reaction quotient. Okay, so we're going to have the concentration of product is 0.6 Molar. And for our reactant it's going to be 0.2 moller and 0.150 molar. So what we're going to get for Q is 20. So we see here that Q is equal to 20 and we want to compare this to K. Okay, so when you compare this to K, we see that Q Que was given in the problem. It's 12.82. So Q is greater than K. And what this tells us is that the reaction is going to proceed in the left direction. Okay. So our reaction is going to shift left to reach equilibrium. So now that we know that we're going to construct an ice table. Okay? And for our ice table, we're going to use this so that we can find out the value that we need to find the concentrations. So the first thing we're going to do is rewrite our reaction. Okay. And for ice table we're going to write ice On the left. And our I wrote is our initial concentration row and were given the initial concentration concentration in the problem. So for sulfur dioxide, our initial concentration is 0.2 for chlorine, it is 0.150. And for sulfuric chloride it is 0.6 00. Okay. And these are all Moeller and we know that the reaction is shifting left, so it's shifting this way. So that means our concentration of products are decreasing and our concentration of reactant are increasing. Okay, so for our change then that means we're going to have minus X. And then we're going to have plus X. And our coefficient is one because we only have one mole of everything. So for R. E. Rowe our equilibrium row of the ice table, we're going to combine both the I and C. Row. So we're going to have 0. plus X. 0.150 plus X. And 0.600 minus x. Okay, So now that we know that we can write out our equilibrium expression R. K. C. Expression based off this ice table and we know that our equilibrium expression is equal to the concentration of products over the concentration of react ints. Okay, so and we know what this value is already. This value is 12.82. So based off of what we have on the ice, the equilibrium row of our ice table, let's set this equal to each other. So our K. is 12 82. And our concentration of products, we're looking at the equilibrium row of our ice table. So this is the information we're going to plug in. So for our products we have 0.600 - X. And for our reactant we have 0.200 plus X Times the concentration of 0.150 plus X. Okay, so that is our equilibrium expression. So now our goal is to solve for X. Okay, so what we're going to do here is we're going to eliminate our our our division. Okay, so we're gonna cross multiply. So what we're going to get then let's go ahead and foil this out. When we foil this part out, What this part becomes is 0.03 plus 0.2 x plus 0.15 X plus X squared. So let's simplify this. And when we simplify it it becomes X squared Plus 0.35 x plus 0.03. Okay, so that's going to be the new denominator. This part here that we foiled out. Alright, so what we're going to get then is when we simplify this out, 12.82 is equal to 0.600 - over. What we just foiled out. Which is X squared Plus 0.35 x plus 0.03. Okay, so we're going to multiply both sides by X squared plus 0.35 X plus 0.3. So what we're gonna get then is 12.82 times X squared plus 0.35 X plus 0. is equal to 0.600 minus X. Okay, so let's foil out the left side. We'll get 12.82 x squared Plus 4.487 x plus 0.3846 is equal to 0.600 -X. Okay, so we want to set both. We want to set this equal to zero. So let's simplify by bringing everything over to one side. When we do that, we get 12.82 X squared plus 4.487 X plus X plus 0. minus 0. equals zero. Okay, And now we have one more step left to simplify which is 12.82, X squared plus 5.487 X -0.2154 is equal to zero. So now what we can do here based off of how this is written is we can use the quadratic formula to solve for X. Because we need to solve for X to plug in X to the equilibrium row of our ice table to solve for the concentration of each species in our reaction. So we need to use the quadratic formula to solve for X. So we'll write that here. Use quadratic formula. Okay, and let's just recall that the quadratic formula is X. Is equal to negative B plus or minus the square root of B squared minus four. A. See over to a. Okay, so let's go ahead and write this out. So this is going to be A. This is B. And this is C. So we're gonna plug in our values into the quadratic formula. So what we're going to get then is X. X is equal to -5. seven plus or minus the square root of B squared which is 5. squared minus four times A. Which is 12 8 2 times C -0. over two times a. Which is 12.82. Okay, So we're going to get X is equal to negative 5.487 plus or minus the square root of 30. Plus 11.045712 over 25.64. Okay, and once we simplify this out we're going to get two values for X. X is equal to 0.36 and X is equal to negative 0.464. So we're going to discard this X. Because We can't have a negative concentration. So we know then that the value for X is going to be the 0.036. So now we can calculate the concentrations at equilibrium because we know this value for X. So let's go ahead and calculate those concentrations. So our concentration for sulfur dioxide based off of the equilibrium table, the equilibrium roast as a 0.200 plus X. So excuse me, that means it's going to be 0.200 plus 0.036. So our concentration of sulfur dioxide Is equal to 0.236 Molar. So that's going to be our first answer. Okay, now we'll soft for our concentration of chlorine and based off the equilibrium role of our ice table, it tells us 0.150 plus X. So that means we have 0.150 plus 0.36. So that gives us a concentration of corinne equal to 0.186 Molar. And lastly our concentration for sulfuric chloride. Okay, so our concentration for sulfuric chloride Is equal to based off the equilibrium of our ice table 0.600 -1. So we have 0. -0.036. Okay, so we're going to have Our concentration of sulfuric chloride is equal to 0.564 Moeller. Okay, so this is going to be our final answer for this problem. These are the concentrations of the species in our reaction. That is it for this problem. I hope this was helpful

Calculate the equilibrium concentrations at 25 °C for the reaction in Problem 15.84 if the initial concentrations are 3N2O44 = 0.0200 M and 3NO24 = 0.0300 M.

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