At 35^∘ C, K=1.6 ×10^-5 for the reaction 2 NOCl(g) ⇌2 NO(g)+Cl2(g) Calculate the concentrations

At 35^∘ C, K=1.6 ×10^-5 for the reaction 2 NOCl(g) ⇌2...

Key Concepts

Reaction Quotient (Q) and Le Châtelier's Principle

The reaction quotient (Q) is calculated in the same way as the equilibrium constant but using the initial concentrations or partial pressures. Comparing Q to K helps determine the direction in which the reaction will proceed to attain equilibrium. Le Châtelier's Principle provides a qualitative understanding of how changes in concentrations, pressure, volume, or temperature will shift the system to counteract those changes and restore equilibrium.

ICE Table Method

The ICE (Initial, Change, Equilibrium) table method is a systematic approach used to organize and calculate the initial concentrations, the changes that occur as the reaction proceeds, and the equilibrium concentrations of all species involved in a reaction. This method is particularly useful for solving equilibrium problems where the equilibrium constant and initial amounts are given.

Equilibrium Constant (K)

The equilibrium constant is a numerical value that expresses the ratio of the concentrations (or pressures) of products to reactants at equilibrium, each raised to the power of their stoichiometric coefficients. It provides insight into the position of the equilibrium and is crucial for quantifying and predicting the outcome of chemical reactions under given conditions.

Chemical Equilibrium

Chemical equilibrium is the state in which both the forward and reverse reactions occur at the same rate, so that the concentrations of reactants and products remain constant over time. This concept is fundamental in understanding how reactions balance out in closed systems and is independent of the fact that the molecular level dynamics continue.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It is essential for setting up the equilibrium constant expression properly and for performing calculations that determine how changes in concentrations affect the system as it approaches equilibrium.

Transcript

00:01 For this problem, we are going to be figuring out the equilibrium concentrations of these species for all of the different conditions that they give us for parts a through f.

00:17 So we're primarily just going to be solving these using ice tables, and that's because they give us some sort of initial concentration.

00:26 We're supposed to be figuring out equilibrium concentrations.

00:29 So we're going to be using ice tables and some of the problems might have multiple ways of solving them and some might have more concepts that i'll talk about later.

00:40 So let's just jump into part a.

00:43 So let's set up our ice table now.

00:48 Let's write down everything that they tell us for our initial conditions.

00:52 So remember, oh, as a side note i want to write on the side.

00:59 They give us a lot of moles and liters.

01:02 So just remember that most of the most of the ones.

01:03 Molarity, which is the measurement for concentration, is equal to moles over volume.

01:12 So that is going to be useful for knowing where i'm getting my numbers from or why i'm putting them that way.

01:22 So i'll put this over here.

01:24 So starting with nocl, initially we have two moles in two liters.

01:32 So we again want to solve for molarity because that is our unit for concentration.

01:36 So we'll do 2 .0 moles in 2 liters.

01:46 So that's going to be 1 .0 molar.

01:52 And then we won't have anything in our products.

01:59 So i'm going to put 0 for both of them.

02:03 Now we're going to do our change.

02:05 So we know that this reaction is going to shift to the right towards the products.

02:13 Because we have zero for the concentrations of the products.

02:17 So we have to make some product in order for it to be in equilibrium, allowing the forward and reverse reaction to both be occurring.

02:25 So for change, that means we're going to be subtracting from our reactants to make products.

02:31 So here we're going to do minus.

02:33 So we have a two in front of our product at ocl, so we're going to do minus 2x.

02:39 The x is because we don't know by how much it's going to change, but it'll change twice.

02:43 As much compared to something that would have a coefficient of one.

02:50 Similarly for nitrogen monoxide, we're changing by 2x, but instead of subtracting, we'll be adding.

02:59 And then for cl2, chlorine gas will be changing by a factor of x.

03:03 And again, we're adding our products.

03:07 And then for our equilibrium concentrations, it's going to be the sum of the previous two rows in the ice table.

03:13 So for nocl, that is going to be 1 .0 minus.

03:18 2x for no it'll be 2x for cl2 it'll be x so now we have our equilibrium concentrations in terms of x we can plug them into an equation for k so k is going to be equal to the concentration of our products so that'll be nitrogen monoxide raised to the power of 2 which is its coefficient times the concentration of cl2 we'll take this we'll divide by the concentration of our reactant, which is n -o -c -l, and we raise that to the power of its coefficient, which is 2.

04:08 We can now plug in the values that we got for our equilibrium concentrations into this equation.

04:14 It's going to be 2x squared.

04:17 Don't forget to square the 2.

04:21 Multiply that by x and divide by 1 .0 minus 2.

04:30 To x.

04:33 Then we can set this equation equal to the value for k, which we were given in the problem as 1 .6 times 10 to the minus 5.

04:42 This allows you to solve for x.

04:44 There are two ways that you can go about that.

04:47 You can use small x approximation, or you can solve it regularly using algebra.

05:09 Alg method is just going to stand for algebraic method, which is just solving the standard equation.

05:15 If you choose to the small x approximately, what you would end up doing is you would end up ignoring this 2x here because what the small x approximation states is that we're going to assume that x will not really change any existing equilibrium concentrations because it's so small.

05:35 But if you choose to do this the small x approximation, make sure to double check because it doesn't always work.

05:45 So you can use this because our k value is less than 10 to the negative 4.

05:57 However, the small x approximation doesn't always work.

06:02 So i prefer to just do it algebraically.

06:05 So again, when we're doing it algebraically, we just include the 2x and then you solve it.

06:11 I use a calculator usually.

06:15 If you write it out by hand you might end up having to factor in certain ways but i use a graphing calculator to solve these so what you end up getting for x either way you use it since the small x approximation works for this one it's not going to work for everything so again you always want to double check for x we are going to get 0 .016 so now we are going to take this x value and we are going to plug it into the equilibrium concentrations that we got in our ice table so that we can solve for the actual values.

07:03 So for the concentration of n -o -c -l, we will get, looking back at our table, it'll be 1 .0 minus 2x where x is .016, and that is going to give us 0 .97 multiple.

07:29 We'll do the same thing for nitrogen monoxide.

07:35 So here, the concentration of nitrogen monoxide is going to be equal to 2x, which is 2 times 0 .016, which is going to be 0 .032.

07:57 And then finally for chlorine gas, it's just equal to x.

08:04 So it's going to be 0 .016 molar.

08:07 And i forgot to write the unit for our other product.

08:12 So they're both molar.

08:14 So these are the answers to part a.

08:18 So we're going to be doing basically the same thing for every single one of these problems.

08:28 So let's just jump into part 2, or part b rather.

08:32 So again, we'll set up our ice table.

08:37 For the reactant, we have zero.

08:40 For our initial, for our products, we are going to have for no, it'll be 2 .0 moles in 1 liter, which is 2 molar.

09:01 And then for chlorine gas, it'll be 1 mole in 1 liter, which is going to be 1 molar.

09:17 In this case, we have 0 for our reactants.

09:20 We want to shift towards our reactants, which is shifting to the left.

09:25 So we're going to have the same changes as the previous problem, except this time we'll switch the signs.

09:35 So for n -o -c -l, instead of subtracting 2x, we're going to add 2x.

09:41 For n -o, we're going to subtract 2x.

09:48 For c -l -2, we will subtract x.

09:52 We can now write in our equilibrium concentrations.

09:56 So it'll be 2x for n -o -cl.

09:59 For no, it'll be 2 .0 minus 2x.

10:05 For cl2, it'll be 1 .0 minus x.

10:11 We can now plug into our k equation.

10:15 So let's just plug in directly, so it'll be 2 .0 minus 2x all of this squared, and this is a concentration, times 1 .0 minus x, we'll divide by 2x, to x and then we get 1 .6 times 10 to the minus 5.

10:50 Again we solve for x and i don't think that the small x approximation is going to work here but for x we're going to get 0 .98 well this shows that the small x approximation won't work because x is a pretty big number whereas previously it was relatively small yeah it was 0 .016.

11:14 So going back to part b.

11:18 Now that we have x we can just plug in to our original equilibrium concentrations that were in terms of x and now we can solve for the actual so n -o -c -l which is equal to 2x at equilibrium which is 2 times .98 so that's going to be about 2 molar for nitrogen monoxide it'll be 2 .0.

11:52 0 minus 2 times .98 which is going to give us the equilibrium concentration which is 0 .040 molar and then finally for c l2 it will be 1 .0 minus 0 .98 so the equilibrium concentration of chlorine gas is going to be equal to 0 .0 to 0 .0 molar.

12:25 So here is the answer to part b.

12:38 So let's move on to part c.

12:48 So for part c, we want to set up our ice table again.

13:00 This time we are going to have, i'm going to give myself a little bit more room, my initial.

13:08 We are going to have one mole of nocl in one liter initially, which is one molar.

13:23 And then for nitrogen monoxide, it'll be the same thing, so it'll also be one molar.

13:32 For cl2, our initial is zero.

13:35 Since we have zero of one of our products, we are going to shift right.

13:44 Remember, when we're shifting right towards the products, we subtract from our reactants and add to our products.

13:49 So for nocl, just like our previous problems, it'll be 2x, and we're subtracting here.

13:55 For no, it'll be plus 2x.

13:59 For cl2, it'll be plus x.