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# hey ...Kc calculation help needed :D watch

1. Hi !!! ....Could anyone give a brief explaination on how to do this question in the attachment! Thanks as ia m clueless to how to do it !
Attached Images

2. The equation for Kc is so the equation will be in this instance

You then sub in the values and calculate for the value of Kc

You can omit the multiplication signs if you want but its just to show you for now.
3. (Original post by Loz17)
The equation for Kc is so the equation will be in this instance

You then sub in the values and calculate for the value of Kc

You can omit the multiplication signs if you want but its just to show you for now.
thanks but ii am aware of this step i am just confused at finding the moles at star and then moles at equlibrium and then finally the concentrations at equilibrium. If you could clarify further on that it would make everything easy then!
4. (Original post by bluefire)
thanks but ii am aware of this step i am just confused at finding the moles at star and then moles at equlibrium and then finally the concentrations at equilibrium. If you could clarify further on that it would make everything easy then!
All the question in the attachement says is find Kc
5. to find Kc you first need to find concentrations of the reactants and products as there are different numbers of moles on both sides of the equations. my method is to write out the equation with mole values at the start and at equilibrium underneath...
.........................4HCl + O2 -> 2Cl2 + 2H2O
start moles...........0.8....0.2...... .0.........0
equilibrium moles....0.2

from this you know that if you've used .6 moles in a 2:1 reaction, the equilibrium moles for each product is 0.3. you also know that for every 4 moles of HCL reacting, 1 mole of O2 reacts, so if 0.6 HCL reacts, so does 0.15 of O2, leaving you with 0.05 moles of O2

.........................4HCl + O2 -> 2Cl2 + 2H2O
start moles...........0.8....0.2...... .0........0
equilibrium moles....0.2...0.05.....0.3..... 0.3

now you can use C=n/V to work out concentrations, since its 10dm^3 divide each by 10, then feed them into the eqation above and out pops the answer
6. above post is correct.
7. (Original post by Artymess)
to find Kc you first need to find concentrations of the reactants and products as there are different numbers of moles on both sides of the equations. my method is to write out the equation with mole values at the start and at equilibrium underneath...
.........................4HCl + O2 -> 2Cl2 + 2H2O
start moles...........0.8....0.2...... .0.........0
equilibrium moles....0.2

from this you know that if you've used .6 moles in a 2:1 reaction, the equilibrium moles for each product is 0.3. you also know that for every 4 moles of HCL reacting, 1 mole of O2 reacts, so if 0.6 HCL reacts, so does 0.15 of O2, leaving you with 0.05 moles of O2

.........................4HCl + O2 -> 2Cl2 + 2H2O
start moles...........0.8....0.2...... .0........0
equilibrium moles....0.2...0.05.....0.3..... 0.3

now you can use C=n/V to work out concentrations, since its 10dm^3 divide each by 10, then feed them into the eqation above and out pops the answer
how did you get the equilibrium moles?? .. iam just confused over here now!!

and yeah thanks alot !! XD
8. (Original post by Artymess)
to find Kc you first need to find concentrations of the reactants and products as there are different numbers of moles on both sides of the equations. my method is to write out the equation with mole values at the start and at equilibrium underneath...
.........................4HCl + O2 -> 2Cl2 + 2H2O
start moles...........0.8....0.2...... .0.........0
equilibrium moles....0.2

from this you know that if you've used .6 moles in a 2:1 reaction, the equilibrium moles for each product is 0.3. you also know that for every 4 moles of HCL reacting, 1 mole of O2 reacts, so if 0.6 HCL reacts, so does 0.15 of O2, leaving you with 0.05 moles of O2

.........................4HCl + O2 -> 2Cl2 + 2H2O
start moles...........0.8....0.2...... .0........0
equilibrium moles....0.2...0.05.....0.3..... 0.3

now you can use C=n/V to work out concentrations, since its 10dm^3 divide each by 10, then feed them into the eqation above and out pops the answer
yeah thats right. i did the same thing. with a different method though
9. equilibrium moles is given away by the question, as they say at equilibrium there are only 0.2 moles of HCl remaining, this means that 0.6 of the original 0.8 have been used up, from that you look at mole ratios, 4 moles of HCL makes 2 moles of each product (2:1 ratio) so you'll get half as many moles of product, for each mole of HCL put in, since you've used 0.6 moles you'll get 0.3 moles of each product at equilibrium.
For the O2 you know that 4 moles of HCl reacts with each mole of O2, hence if you used 0.6 moles, then a quarter of it is the amount of O2 used, which is 0.15. you started with 0.2 moles of O2, you used 0.15, hence you have 0.05 remaining at equilibrium
10. (Original post by Artymess)
equilibrium moles is given away by the question, as they say at equilibrium there are only 0.2 moles of HCl remaining, this means that 0.6 of the original 0.8 have been used up, from that you look at mole ratios, 4 moles of HCL makes 2 moles of each product (2:1 ratio) so you'll get half as many moles of product, for each mole of HCL put in, since you've used 0.6 moles you'll get 0.3 moles of each product at equilibrium.
For the O2 you know that 4 moles of HCl reacts with each mole of O2, hence if you used 0.6 moles, then a quarter of it is the amount of O2 used, which is 0.15. you started with 0.2 moles of O2, you used 0.15, hence you have 0.05 remaining at equilibrium
i do appreciate your method but don't you think alot of oral is involved in it how come its not like this that if .200 mol of hcl remained as it is of 4:2 ratio so .100 mol of cl2 and h2o will remain???
11. (Original post by kosy91)
yeah thats right. i did the same thing. with a different method though
sorry to bother you again can you tell me your method as well!! thanks
12. (Original post by bluefire)
i do appreciate your method but don't you think alot of oral is involved in it how come its not like this that if .200 mol of hcl remained as it is of 4:2 ratio so .100 mol of cl2 and h2o will remain???
The idea is that you use reactants to make products, so if you begin with 0.8 moles, and at equilibrium 0.2 moles remain, then the missing 0.6 moles are what have been used to form the products, so its the 0.6 that you apply the ratio to
13. (Original post by Artymess)
The idea is that you use reactants to make products, so if you begin with 0.8 moles, and at equilibrium 0.2 moles remain, then the missing 0.6 moles are what have been used to form the products, so its the 0.6 that you apply the ratio to
understood for now and for equlibrium moles of HCl we will take .2 and not the missing .6 right ? If tahts so then i am done for now!! thanks once again!!
14. thats right, stick with the 0.2 for HCl moles at equilibrium, but use the 0.6 value to work out the remaining missing values
15. (Original post by Artymess)
equilibrium moles is given away by the question, as they say at equilibrium there are only 0.2 moles of HCl remaining, this means that 0.6 of the original 0.8 have been used up, from that you look at mole ratios, 4 moles of HCL makes 2 moles of each product (2:1 ratio) so you'll get half as many moles of product, for each mole of HCL put in, since you've used 0.6 moles you'll get 0.3 moles of each product at equilibrium.
For the O2 you know that 4 moles of HCl reacts with each mole of O2, hence if you used 0.6 moles, then a quarter of it is the amount of O2 used, which is 0.15. you started with 0.2 moles of O2, you used 0.15, hence you have 0.05 remaining at equilibrium
PCl5 decomposes to form PCl3 and Cl2. Given that eqilibrium constant is 11.5 moldm^-3at 3ooK what is the eqilibrium concentration of PCl3 when conc. of PCl5 at equilibrium is 0.07 moldm^-3 ?

I don't have the answer for this one my teacher gave it to us to solve it and submit tomorrow but i got the answer as 0.897 moldm^-3.
16. I just worked it out and got 0.897 moldm^-3
17. (Original post by Artymess)
I just worked it out and got 0.897 moldm^-3
so that means i am right then thanks man lets see what the teacher tells us tomorrow actually these questions are being attempted for the firsst time tha's why all these doubts anyhow have a nice time and take care ..bye!!
18. (Original post by bluefire)
sorry to bother you again can you tell me your method as well!! thanks
thats not a problem.

your method was based on logic. mine is based on calculation.

at start:................ 4(0.2)..........0.2..........0.. ........0
at equilibrium:..........4(0.2-x).......0.2-x.......2x........2x
but; ................4(0.2-x)=0.2
therefore; .............x=0.15
moles at equi:...........0.2...........0. 05.........0.3......0.3
from here you can get conc and hence Kp
so it comes to the same thing as yours.
19. (Original post by Artymess)
I just worked it out and got 0.897 moldm^-3
....that

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