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mole calculations q

for part ii) it wants delta H - so since the HCL is in excess i used the mole ratio from the equation given
Ca(OH)2 + 2HCL --> CaCl2 +2H2O
so 0.001 * 2 gives 0.002 but the markscheme used the number of moles as 0.001 ..why??????
3b.PNG25.0KB
Original post by pondsteps
for part ii) it wants delta H - so since the HCL is in excess i used the mole ratio from the equation given
Ca(OH)2 + 2HCL --> CaCl2 +2H2O
so 0.001 * 2 gives 0.002 but the markscheme used the number of moles as 0.001 ..why??????


and im stuck at another part ..
its still the same number of moles.. so how does it effect temp change ??
(edited 7 years ago)
9.PNG10.0KB
Original post by pondsteps
and im stuck at another part ..
its still the same number of moles.. so how does it effect temp change ??


Temperature change increases, faster rate, because more particles in a smaller given volume, therefore more frequent successful collisions.
Original post by Asad_2015
Temperature change increases, faster rate, because more particles in a smaller given volume, therefore more frequent successful collisions.

yeah but how did u know that temp change increases when no. of moles remain the same???
Original post by pondsteps
yeah but how did u know that temp change increases when no. of moles remain the same???


twice as small volume, some twice as many particles in volume, so doubles. and the enthalpy question, the HCL is in excess so you divide by 0.001 moles as that's how much Cah02 can react only :s
Original post by pondsteps
for part ii) it wants delta H - so since the HCL is in excess i used the mole ratio from the equation given
Ca(OH)2 + 2HCL --> CaCl2 +2H2O
so 0.001 * 2 gives 0.002 but the markscheme used the number of moles as 0.001 ..why??????


SORRRY just realised (facepalm) BASCIALLY :P enthalpy change delta H is standard enthalpy molar change. So you want to know the molar enthalpy change yes you can divide by 0.002 (2 mols) but you have to divide by 2 to get 1 mole. As units are KJ/mol-1 (enthaply change per mole, not 2 moles :P)
Original post by Asad_2015
twice as small volume, some twice as many particles in volume, so doubles.

so as the volume decreases the temp change increases???
Original post by Asad_2015
SORRRY just realised (facepalm) BASCIALLY :P enthalpy change delta H is standard enthalpy molar change. So you want to know the molar enthalpy change yes you can divide by 0.002 (2 mols) but you have to divide by 2 to get 1 mole. As units are KJ/mol-1 (enthaply change per mole, not 2 moles :P)


ohhhh lol - ok thanks x

btw, if the solution was not in excess and we find no. of moles by multiplying conc. with volume we wont need to divide by 2 right?
(edited 7 years ago)
Original post by pondsteps
so as the volume decreases the temp change increases???


I think if you were to half the volume, there wouldn't be a change since the number of moles would be halved and the space where collisions may occur will also be halved. They seem to just cancel out.

However, if you were to half the volume and double the moles per unit volume, there would be an increase in frequency of successful collisions as there's twice as many moles in that given volume (which was halved) so kinetic energy would be accumulated more easily.
Original post by pondsteps
ohhhh lol - ok thanks x

btw, if the solution was not in excess and we find no. of moles by multiplying conc. with volume we wont need to divide by 2 right?


NOO. If the solution is not in excess and the ratio is 1:2 you divide by 1 mol of the substance. Because it's enthalpy change per mole! But say if they never told you what was in excess you'd have to find out, cause if you only have 0.1 moles of oxygen and 0.05 mole of carbon then only there 0.05 mole to react, so you divide by 0.05. I'm confusing you now just read this page :tongue: image.jpg
Original post by M0nkey Thunder
I think if you were to half the volume, there wouldn't be a change since the number of moles would be halved and the space where collisions may occur will also be halved. They seem to just cancel out.

However, if you were to half the volume and double the moles per unit volume, there would be an increase in frequency of successful collisions as there's twice as many moles in that given volume (which was halved) so kinetic energy would be accumulated more easily.

but here the volume decreased and the no. of moles remained the same
Original post by Asad_2015
NOO. If the solution is not in excess and the ratio is 1:2 you divide by 1 mol of the substance. Because it's enthalpy change per mole! But say if they never told you what was in excess you'd have to find out, cause if you only have 0.1 moles of oxygen and 0.05 mole of carbon then only there 0.05 mole to react, so you divide by 0.05. I'm confusing you now just read this page :tongue: image.jpg

ohh i think i got it - thanks
(edited 7 years ago)
Original post by pondsteps
but here the volume decreased and the no. of moles remained the same


But the number of moles per unit volume doubled as the volume halved to keep the number of moles the same. If the the concentration didn't increase by the same factor the volume decreased, the temperature change wouldn't have increased because we wouldn't have the same number of moles any more as n= c * v

So your previous statement would have been fine if you said: for the same number of moles in a given volume, as the volume decreased, the temperature change increased.
Original post by M0nkey Thunder
But the number of moles per unit volume doubled as the volume halved to keep the number of moles the same. If the the concentration didn't increase by the same factor the volume decreased, the temperature change wouldn't have increased because we wouldn't have the same number of moles any more as n= c * v

So your previous statement would have been fine if you said: for the same number of moles in a given volume, as the volume decreased, the temperature change increased.


thank u so much :smile:
Original post by pondsteps
thank u so much :smile:


No worries :smile:
Original post by M0nkey Thunder
I think if you were to half the volume, there wouldn't be a change since the number of moles would be halved and the space where collisions may occur will also be halved. They seem to just cancel out.

However, if you were to half the volume and double the moles per unit volume, there would be an increase in frequency of successful collisions as there's twice as many moles in that given volume (which was halved) so kinetic energy would be accumulated more easily.

quick question here, is the first case for gases and the sec for solids and liquids? (cos the no. of moles can't be randomly halved...
Original post by pondsteps
quick question here, is the first case for gases and the sec for solids and liquids? (cos the no. of moles can't be randomly halved...


The number of moles would be halved if you only halved the volume if you look at the equation: Moles= concentration *volume.

The only way to keep the number of moles the same when you half the volume is by doubling the concentration.

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