15. Consider the reaction: 3 H2(g) + N2(g) 2 NH3(g) The...

15. Consider the reaction: 3 H$_2$(g) + N$_2$(g) \iff 2 NH$_3$(g) The following gases are placed in a sealed flask at initial partial pressures of: H$_2$(g) 10 atm N$_2$(g) 30 atm NH$_3$(g) 20 atm After equilibration at 25°C, the equilibrium partial pressure of H$_2$ is 16 atm. What is the equilibrium partial pressure of NH$_3$ in atmospheres? (a) The value of the equilibrium constant is required to solve this problem. (b) 24 (c) 16 (d) 14 (e) 11

Transcript

00:01 For this problem, we're doing a couple of different ice table problems and determining some information.

00:05 For the first part, we were given equilibrium concentrations and a kc value and asked to find the concentration of one of the reactants.

00:15 So our kc expression is our products raised to their coefficient.

00:21 So nh3 squared over the reactants raised to their coefficients.

00:27 And so then we would have h2 cubed.

00:30 So we are told that kc is 64.

00:33 We know our nh3 is .350, so we'll go ahead and square that.

00:39 And we know that our n2 is .0192, and then the rest of this will be our h2 cubed.

00:46 So let's go ahead and solve for what h2 cubed would be.

00:52 So we would have .350 squared divided by 64 and divided by .0192, and that equals .0997.

01:11 And then when we take the cubed root of that to find our concentration of h2, it's going to be .464, and it is a concentration, so we'll make that molar.

01:23 So that is what would go right here.

01:27 Okay, so that's the first problem.

01:29 The second problem has a couple different parts to it.

01:33 And they give us initial concentrations, and then they gave us an equilibrium concentration for the no2.

01:40 So that tells you that this actually has the minus, because it was reduced, this is the minus 2x.

01:47 Again, it's minus 2x because that's a 2.

01:50 This will be plus x.

01:52 So whatever we determine x to be, then we will simply add that to this 1 .580.

01:59 So because we know that this went from .85 to .512, we need to find the difference between those, and that equals .338, and that's 2x.

02:17 So then x is going to be half of that, which is going to be 0 .169.

02:23 So we now know that this is 0 .169 that we are going to add onto this.

02:30 So plus 1 .580, and that would give us our equilibrium concentration of 1 .749.

02:41 Oh, this was pressure, not concentration, atmospheres.

02:47 And then they're asking us to find kp...

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