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.
5H.2 Clarifications
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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
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
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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.
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
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