## 5H.2 Clarifications

Phuong Tran 1G
Posts: 33
Joined: Sat Sep 14, 2019 12:16 am

### 5H.2 Clarifications

Hi! So I’m reviewing and 5h.2 states hat chemical equations can be expressed as sum of 2 or more chemical equations. The book gave 3 gas phase equations and did the following:

2 P(g) + 3 Cl2(g) <—-> 2 PCl3(g)
2 PCl3(g) + 2 Cl2(g) <—-> 2 PCl5(g)
2 P(g) + 5 Cl2(g) <—-> 2 PCl5(g)

So they added the two equations and got the third one (last row). I’m confused on how they added it? I know 3+2 CL2 would give 5.

Brandi 2C
Posts: 52
Joined: Thu Feb 28, 2019 12:15 am

### Re: 5H.2 Clarifications

They simply added the 1st and 2nd equations together and kept the molecules on their respective sides.

2 P(g) + 3 Cl2(g) <—-> 2 PCl3(g)
+ 2 PCl3(g) + 2 Cl2(g) <—-> 2 PCl5(g)
= 2 P + 3 Cl2 + 2 PCl3 + 2 Cl2 <-> 2 PCl3 + 2 PCl5

and you could subtract 2 PCl3 from each side and therefore removing them, and as you said add the Cl2's together which would give the third equation you described

2 P(g) + 5 Cl2(g) <—-> 2 PCl5(g)

Hope this helps. I feel like you understood the concept of adding the Cl2 together but didn't notice that the PCl3 would get removed from the final equation

Phuong Tran 1G
Posts: 33
Joined: Sat Sep 14, 2019 12:16 am

### Re: 5H.2 Clarifications

Brandi 2C wrote:They simply added the 1st and 2nd equations together and kept the molecules on their respective sides.

2 P(g) + 3 Cl2(g) <—-> 2 PCl3(g)
+ 2 PCl3(g) + 2 Cl2(g) <—-> 2 PCl5(g)
= 2 P + 3 Cl2 + 2 PCl3 + 2 Cl2 <-> 2 PCl3 + 2 PCl5

and you could subtract 2 PCl3 from each side and therefore removing them, and as you said add the Cl2's together which would give the third equation you described

2 P(g) + 5 Cl2(g) <—-> 2 PCl5(g)

Hope this helps. I feel like you understood the concept of adding the Cl2 together but didn't notice that the PCl3 would get removed from the final equation

Yes! Thank you very much.

Brandi 2C
Posts: 52
Joined: Thu Feb 28, 2019 12:15 am

### Re: 5H.2 Clarifications

Phuong Tran 1G wrote:
Brandi 2C wrote:They simply added the 1st and 2nd equations together and kept the molecules on their respective sides.

2 P(g) + 3 Cl2(g) <—-> 2 PCl3(g)
+ 2 PCl3(g) + 2 Cl2(g) <—-> 2 PCl5(g)
= 2 P + 3 Cl2 + 2 PCl3 + 2 Cl2 <-> 2 PCl3 + 2 PCl5

and you could subtract 2 PCl3 from each side and therefore removing them, and as you said add the Cl2's together which would give the third equation you described

2 P(g) + 5 Cl2(g) <—-> 2 PCl5(g)

Hope this helps. I feel like you understood the concept of adding the Cl2 together but didn't notice that the PCl3 would get removed from the final equation

Yes! Thank you very much.

No problem! Good luck studying and considering how quickly you've replied you might already be haha