## 5I.5

smurphy1D
Posts: 51
Joined: Mon Nov 18, 2019 12:18 am

### 5I.5

In this equation I am given the value of K and 2 partial pressure values. I need to solve to find the P for the last one. I am dividing and multiplying the other values around to solve for x but I keep getting the wrong answer. Am I forgetting a step?
Ppcl5= Ppcl3 + Pcl2
1.8 bar =(x) + 5.43 bar K=25 25=5.43(x)/1.8. 25 x 1.8/5.43=x

Renee Grange 1I
Posts: 56
Joined: Fri Aug 30, 2019 12:16 am
Been upvoted: 1 time

### Re: 5I.5

I think the reason you are getting the wrong answer is because you are using 1.8 bar instead of 1.18 bar, but your steps are correct.

Posts: 81
Joined: Fri Sep 28, 2018 12:28 am

### Re: 5I.5

You did everything right, change 1.8 to 1.18 and you will get the correct outputs.

Pegah Nasseri 1K
Posts: 100
Joined: Wed Feb 27, 2019 12:15 am

### Re: 5I.5

For this problem you know that K=25 so using the chemical equation $PCl_{5} (g) \rightleftharpoons PCl_{3} (g) + Cl_{2} (g)$ you know that:
$K = \frac{(P_{PCl_{3}})(P_{Cl_{2})}}{P_{PCl_{5}}} = 25$
Plug in the values of the partial pressures that have already been given to you:
$K= \frac{P_{PCl_{3}}(5.43)}{1.18}=25$
Then isolate PPCl3 to find its partial pressure:
$P_{PCl_{3}}= \frac{(25)(1.18)}{5.43}= 5.4 bar$

smurphy1D
Posts: 51
Joined: Mon Nov 18, 2019 12:18 am

### Re: 5I.5

Renee Grange 1I wrote:I think the reason you are getting the wrong answer is because you are using 1.8 bar instead of 1.18 bar, but your steps are correct.

Yes, got it thank you!