Sulfur dioxide reacts with nitrogen dioxide forming sulfur...

Sulfur dioxide reacts with nitrogen dioxide, forming sulfur trioxide and nitrogen monoxide. a. Write a balanced chemical reaction, all are gases. b. If the value of Kc for the reaction is 2.50, what are the equilibrium concentrations of the products if the initial amounts of each are 0.50 M SO2, 0.50 M NO2, 0.0050 M SO3, and 0.0050 M NO.

Transcript

00:01 All right, so we're told that the equilibrium content for this reaction is equal to 224.

00:09 And we are given some initial concentrations, and we want to know what's the concentration of so3 at equilibrium.

00:14 So let's set up an ice chart.

00:17 So initially, we have six moles each of sulfur dioxide and nitrogen dioxide, so these two, in two liters.

00:25 So six divided by two is 3 .0 molar.

00:28 We always want the concentrations in molar.

00:33 And we have zero of this because it doesn't say anything about them, which means since we have only reactants to start, they're going to decrease in concentration.

00:42 This is, yeah, this is balanced.

00:44 So it's all one to one.

00:46 And these are going to increase in concentration.

00:48 You can only go in one possible way because there's no way to, you know, transform non -existent product into more reactive.

00:55 So at equilibrium, this is three minus x.

00:59 This is three minus x.

01:01 This is x.

01:02 And this is x and the k value is going to be the concentration of each of these products so x times x divided by 3 minus x times 3 minus x the reactants so this is x squared over 3 minus x squared and it's going to be equal to 2 24 luckily because we have squared on both top and bottom we can take the square root and then we'll get x over 3 minus x is equal to the square root of 24.

01:44 So the square root of 24 is going to be plus or minus 14 .97.

01:56 All right.

01:57 So x is equal to multiply both sides by 3 minus x.

02:04 I'll take the positive one first...

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