At T = 25°C, the reaction 2 NO2 (g) 2 NO (g) + O2(g) has an...

At T = 25°C, the reaction 2 NO2 (g) \iff 2 NO (g) + O2(g) has an equilibrium constant Kp = 5.0 × 10^-13. Suppose the container is filled with NO2 at an initial pressure of 0.25 atm. Calculate the partial pressure of NO at equilibrium. Consider the following two equilibria occurring simultaneously in solution. HIn (aq) \iff H^+ (aq) + In^- (aq) methyl orange: red yellow CH3COOH (aq) \iff H^+ (aq) + CH3COO^- (aq) (1) (2)

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00:01 So we're given the equilibrium a plus 2b yields 3c, and we're told that the kc is 0 .25.

00:10 We're asked which diagram represents the fact that we are at equilibrium.

00:17 And if the system is not at equilibrium, predict which direction it would go.

00:22 Understanding that each molecule represents 0 .4 moles, and the volume of the container in each case is 2 liters.

00:36 So when we look at a, we can look at the red.

00:43 We have three of them.

00:45 The blue, we have three of them.

00:50 And the green, we have two of them.

00:55 So that means that we're going to multiply them all by 0 .4, and then we're going to divide it by 2 liters.

01:08 So for the first one, we get a concentration of a is 3 times 0 .4 divided by 2 gives us 0 .6.

01:23 So that's 0 .6 molar, 0 .6 molar.

01:33 And then we have 2 times 0 .4 divided by 2 is going to give us 0 .4 molar.

01:41 For b, we have for the red, we have 4.

01:48 For the blue, we have 2.

01:51 And for the green, we have 2.

01:55 So once again, that's going to be times 0 .4 divided by 2 times 0 .4 divided by 2 times 0 .4 divided by 2.

02:09 So that's going to be 4 times 0 .4 divided by 2 is going to give me 0 .8.

02:18 So that's going to be 0 .8 molar.

02:21 This is going to be 0 .4 molar.

02:24 And this is going to be 0 .4 molar.

02:27 And then for the last one, we have for our red, we have 3, which i already know that 3 is going to give me 0 .6 molar.

02:40 For the green, we have 4, and that's going to give me 0 .8 molar.

02:51 And for the blue, i have 1, which is going to give me 0 .2 molar.

03:04 So now what we have to do is for each of the situations, we have to figure out what q is equal to.

03:10 So since k, my kc is going to be equal to the concentration of c cubed divided by the concentration of a times the concentration of b squared.

03:25 And i know that that's equal to 0 .25.

03:28 For each situation, i have to calculate what q is.

03:33 So for qc for a, i get, i actually have to remember which is which...

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