# Chemistry IGCSE energetics

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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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#2

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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).

**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).

What are your thoughts? What have you tried so far?

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#4

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So I don't know if it would be doubled or halved for the temperature. I'm sorry, I am quite stuck.

**PandagirlRLC**)So I don't know if it would be doubled or halved for the temperature. I'm sorry, I am quite stuck.

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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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...

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

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.

**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

(Original post by

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...

**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...

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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(Original post by

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!!!

**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!!!

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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(Original post by

I got 23.3

**ALikesGeetars**)I got 23.3

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#10

(Original post by

How did you work it out?

**PandagirlRLC**)How did you work it out?

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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#11

(Original post by

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???

**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???

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 /

__= - 146.3 J/mol" (NOT 75)__

**50**
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#12

(Original post by

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 /

**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 /

__= - 146.3 J/mol" (NOT 75)__**50**
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**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 /

__= - 146.3 J/mol" (NOT 75)__

**50**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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#14

(Original post by

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?

**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?

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