# Chemistry IGCSE energetics

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#1
When 25 cm3 of a solution of sodium hydroxide was mixed with 25 cm3 of hydrochloric acid of the same concentration the temperature change was found to be 35oC.*What would the temperature change be if 25 cm3 of the same solution of sodium hydroxide was mixed with 50 cm3 of the same hydrochloric acid? Explain your answer. (Think carefully about moles reacting and what you know about measuring energy changes).
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6 days ago
#2
(Original post by PandagirlRLC)
When 25 cm3 of a solution of sodium hydroxide was mixed with 25 cm3 of hydrochloric acid of the same concentration the temperature change was found to be 35oC.*What would the temperature change be if 25 cm3 of the same solution of sodium hydroxide was mixed with 50 cm3 of the same hydrochloric acid? Explain your answer. (Think carefully about moles reacting and what you know about measuring energy changes).
Nice copy and pasta.

What are your thoughts? What have you tried so far?
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#3
(Original post by Pigster)
Nice copy and pasta.

What are your thoughts? What have you tried so far?
So I don't know if it would be doubled or halved for the temperature. I'm sorry, I am quite stuck.
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6 days ago
#4
(Original post by PandagirlRLC)
So I don't know if it would be doubled or halved for the temperature. I'm sorry, I am quite stuck.
Neither doubled, nor halved.

Do q = mcDT and work out DH (this value is a constant), then work it back with the new value for m to work out DT.
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#5
so DH = q/n
and then it would normally be q = 50 x 4.18 x 35 = 7315 J
and n=mass/Mr and so n for NaOH = 25/(23+16+1) which is 0.625 and then n for HCl = 25/(1+35.5)=50/73
so total n would be 0.625 + 50/73 = 765/584 or 1.3
and so DH would be 7315/765/584 = 5584

with the new amounts:
q = 75 x 4.18 x DT
DH = q/n
n for NaOh remains 0.625 but n for HCl = 50/36.5 = 100/73
so total n would be 0.625 + 100/73 = 1.99
and so DH would be 313.5 x DT / 1.99 = 5584
so 313.5 x DT = 11112.16
and then DT = 35???

I think I have gone wrong...

(Original post by Pigster)
Neither doubled, nor halved.

Do q = mcDT and work out DH (this value is a constant), then work it back with the new value for m to work out DT.
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6 days ago
#6
(Original post by PandagirlRLC)
so DH = q/n
and then it would normally be q = 50 x 4.18 x 35 = 7315 J
and n=mass/Mr and so n for NaOH = 25/(23+16+1) which is 0.625 and then n for HCl = 25/(1+35.5)=50/73
so total n would be 0.625 + 50/73 = 765/584 or 1.3
and so DH would be 7315/765/584 = 5584

with the new amounts:
q = 75 x 4.18 x DT
DH = q/n
n for NaOh remains 0.625 but n for HCl = 50/36.5 = 100/73
so total n would be 0.625 + 100/73 = 1.99
and so DH would be 313.5 x DT / 1.99 = 5584
so 313.5 x DT = 11112.16
and then DT = 35???

I think I have gone wrong...
The value for n you should be using is the amount of water formed (check your definition for enthalpy of neutralisation).

Try again, a cursory check looks like your method is fine, otherwise.

Don't forget your minus sign for DH for an exo. reaction!!!
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#7
(Original post by Pigster)
The value for n you should be using is the amount of water formed (check your definition for enthalpy of neutralisation).

Try again, a cursory check looks like your method is fine, otherwise.

Don't forget your minus sign for DH for an exo. reaction!!!
Thank you.
I thought n was number of moles?
so would it just be 50 for the first one (25+25) and 75 for the changed one (25+50)?

so DH = q/n
and then it would normally be q = 50 x 4.18 x 35 = 7315 J
and so DH would be 7315/50 = - 146.3 J/mol

with the new amounts:
q = 75 x 4.18 x DT
DH = q/n
and so DH would be 313.5 x DT / 75 = - 146.3 J/mol
so 313.5 x DT = 10972.5 J
and then DT = 35 ???

why do I keep on getting 35???
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6 days ago
#8
I got 23.3
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#9
(Original post by ALikesGeetars)
I got 23.3
How did you work it out?
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5 days ago
#10
(Original post by PandagirlRLC)
How did you work it out?
I may be wrong, it's been a while since I did IGCSE Chemistry

Firstly let's remember Q = MCDT

25cm3 NaOH and 25cm3 HCL = 50cm3 volume and DT = 35

Q= 50(M) x 4.18(C) x 35(DT)

= 7315 J

Now we want to find the new temperature with a new volume of NaOH and HCl

25 + 50 = 75cm3

7315 = 75 x 4.18 x X(DT)

X = 7315 / 313.5(which is 75x4.18)

= 23.33333
=23.3

I hope I did it right, as I've said it's been a while hehe
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5 days ago
#11
(Original post by PandagirlRLC)
Thank you.
I thought n was number of moles?
so would it just be 50 for the first one (25+25) and 75 for the changed one (25+50)?

why do I keep on getting 35???
n(water) is the same for both.

In the first case n(water) = n(acid) = n(alkali) since neither is in excess.

In the second case n(acid) > n(alkali), so n(water) = n(alkali) NOT n(both combined).

So...

"with the new amounts:
...
and so DH would be 313.5 x DT / 50 = - 146.3 J/mol" (NOT 75)
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5 days ago
#12
(Original post by Pigster)
n(water) is the same for both.

In the first case n(water) = n(acid) = n(alkali) since neither is in excess.

In the second case n(acid) > n(alkali), so n(water) = n(alkali) NOT n(both combined).

So...

"with the new amounts:
...
and so DH would be 313.5 x DT / 50 = - 146.3 J/mol" (NOT 75)
Ahhh that makes much more sense!
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#13
(Original post by Pigster)
n(water) is the same for both.

In the first case n(water) = n(acid) = n(alkali) since neither is in excess.

In the second case n(acid) > n(alkali), so n(water) = n(alkali) NOT n(both combined).

So...

"with the new amounts:
...
and so DH would be 313.5 x DT / 50 = - 146.3 J/mol" (NOT 75)
Thank you so much, so would the answer be
313.5 x DT = -7315
DT = -23
so DT = 23 C?

and then my explanation would be that there are now more moles of acid reacting whereas before neither were in excess. the energy change always stays the same and so you rearrange to find DT as 23?
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3 days ago
#14
(Original post by PandagirlRLC)
Thank you so much, so would the answer be
313.5 x DT = -7315
DT = -23
so DT = 23 C?

and then my explanation would be that there are now more moles of acid reacting whereas before neither were in excess. the energy change always stays the same and so you rearrange to find DT as 23?
In the second mixture, there is still the same number of successful collisions (as the excess acid cannot react), so there is the same amount of energy released (since each collision releases the same amount of energy), BUT this energy is spreading out into a larger volume of mixture. In fact there is 3/2x more mixture, so the heating effect is 2/3x DT, i.e. 2/3x 35 = 23.
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