Consider the reaction: A reaction mixture at equilibrium at...

Name: Problem 81, Chapter 15: Chemistry: Structure and Properties
Uploaded: 2019-12-10T03:59:18-08:00
Duration: 6 min 44 s
Channel: William Mills
Description: Consider the reaction: A reaction mixture at equilibrium at 175 K contains PH2 = 0.958 atm, PI2 = 0.877 atm, and PHI = 0.020 atm. A second reaction mixture, also at 175 K, contains PH2 = PI2 = 0.621 atm, and PHI = 0.101 atm. Is the second reaction at equilibrium? If not, what will be the partial pressure of HI when the reaction reaches equilibrium at 175 K?

Consider the reaction: A reaction mixture at equilibrium at 175 K contains PH2 = 0.958 atm, PI2 = 0.877 atm, and PHI = 0.020 atm. A second reaction mixture, also at 175 K, contains PH2 = PI2 = 0.621 atm, and PHI = 0.101 atm. Is the second reaction at equilibrium? If not, what will be the partial pressure of HI when the reaction reaches equilibrium at 175 K?

Consider the reaction:
A reaction mixture at equilibrium at 175 K contains PH2 = 0.958 atm,
PI2 = 0.877 atm, and PHI = 0.020 atm. A second reaction
mixture, also at 175 K, contains PH2 = PI2 = 0.621 atm, and
PHI = 0.101 atm. Is the second reaction at equilibrium? If not,
what will be the partial pressure of HI when the reaction reaches
equilibrium at 175 K?

Transcript

00:01 This question asks us to consider a reaction where h2 combines with i2 to form two moles of hi.

00:08 It tells us that the mixture of equilibrium has the following pressures of each of the gases, and that it tells us there's a second mixture that has different pressures for each of the gases.

00:23 And the first question asks is, is the second reaction at equilibrium? so above this horizontal line, i have what it's, says are the equilibrium conditions.

00:33 It says at equilibrium at 175 kelvin, the pressure of h2 is 0 .958 atmospheres, the pressure of i2 is 0 .877 atmospheres, and the pressure of h .i is 0 .02 atmospheres.

00:47 Here's the equation for the equilibrium constant with respect to pressure, kp.

00:52 It's the partial pressure due to hi squared because there's a two here, divided by the product of the partial pressure of h2 and the partial pressure of i2.

01:03 If you plug in each of the numbers you gave us, 0 .02 for hi, and 0 .958, and 0 .877 for h2 and i2 respectively, you would see that the equilibrium constant with respect to pressure of the equilibrium, that equilibrium, is 4 .76 times 10 to the negative 4th.

01:21 So that's one bit of information.

01:23 Then it asks is the second reaction in equilibrium.

01:27 And it tells us the pressures of each of the gases under these conditions.

01:30 It tells us the h2 is 0 .621, and i2 is also 0 .621, and h .i is 0 .101 atmospheres.

01:38 If we plug these values into the same equation to calculate partial pressure with respect to p or kp, we would see that it's equal room consequence.

01:48 This would be the reaction quotient or q really at this point.

01:51 So i'm going to write this as a q is 2 .65 times 10 to the negative.

01:57 Second we can tell that these two numbers of course are not the same and this is actually much larger than the equilibrium constant of respect to pressure so when q is greater than kp you know that the reaction is going to run backwards i'll say that again when q or the reaction quotient is larger than kp it favors the reactants and the reaction run to the left or backwards as it's written here and so that is the first question no it's not equilibrium but now let's start know once it does get to equilibrium, what will the partial pressure of h .i.

02:32 Actually be.

02:34 So i said at the beginning of an ice table here.

02:36 Let's finish it.

02:39 So i said that the reaction is running backwards.

02:41 So normally when we write it up, we write minus for each of the reactants.

02:45 But since we know it's moving to the left, these changes will be positive for the reactants and they will be negative for the products.

02:58 Because we're going to go backwards and take h -i in form h2 and i -2, we're going to lose h -i, and we're going to gain h -2 and i -2.

03:07 Additionally, we'll lose twice as much h -i as we will gain for any one of the two reactants because of this two here...

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