Chemsitry Enthalpy Change Question Help

Watch
Principleturn
Badges: 4
Rep:
?
#1
Report Thread starter 1 year ago
#1
When ethanoic acid reacts with sodium hydroxide, the enthalpy change, ∆H, is
–56.1 kJ mol–1

CH3COOH(aq) + NaOH(aq) ⟶ CH3COONa(aq) + H2O(l)

Calculate the temperature rise when 25 cm3 of 2.0 mol dm–3 aqueous ethanoic acid
react with 25 cm3 of 2.0 mol dm–3 aqueous sodium hydroxide.

Assume that both solutions have the same initial temperature, have a density of
1.0 g cm–3 and a specific heat capacity of 4.18 J K–1 g–1

How would you do this? Obviously q=mcT

But would you work out moles, set out 1:1 ratio then use that as a mass of water? Help !
0
reply
Kian Stevens
Badges: 16
Rep:
?
#2
Report 1 year ago
#2
The moles of both ethanoic acid and sodium hydroxide are \frac{2*25}{1000} = 0.05 mol. Due to 1:1 stoichiometry, this means that 0.05 mol of water are formed.
Since \Delta H = \frac{Q}{n}, this means that Q = \Delta H n. Hence, Q = 56.1 * 0.05 = 2.805 kJ = 2805 J (the reason why I dropped the negative sign in the enthalpy change is because it simply shows that the enthalpy change is exothermic. You can't have negative energy, which would arise if I was to have kept the negative sign).

Now we have to use Q=mc \Delta T. Rearranging for \Delta T would give \Delta T = \frac{Q}{mc}. Let's make a checklist of the terms in the equation:
  • We calculated Q above; it's 2805 J.
  • c is 4.18 J g-1 K-1.
  • m is 50 g. Since it's assumed that both solutions have a density of 1 g cm-1, and both solutions have volumes of 25 cm3, this means they both have masses of 25 g. So, mixing these two solutions would give a combined volume of 50 cm3, and so a combined mass of 50 g.

So, we have everything to calculate the temperature change. \Delta T = \frac{2805}{50*4.18} = 13.4 K.
Last edited by Kian Stevens; 1 year ago
1
reply
Principleturn
Badges: 4
Rep:
?
#3
Report Thread starter 1 year ago
#3
(Original post by Kian Stevens)
The moles of both ethanoic acid and sodium hydroxide are \frac{2*25}{1000} = 0.05 mol. Due to 1:1 stoichiometry, this means that 0.05 mol of water are formed.
Since \Delta H = \frac{Q}{n}, this means that Q = \Delta H n. Hence, Q = 56.1 * 0.05 = 2.805 kJ = 2805 J (the reason why I dropped the negative sign in the enthalpy change is because it simply shows that the enthalpy change is exothermic. You can't have negative energy, which would arise if I was to have kept the negative sign).

Now we have to use Q=mc \Delta T. Rearranging for \Delta T would give \Delta T = \frac{Q}{mc}. Let's make a checklist of the terms in the equation:
  • We calculated Q above; it's 2805 J.
  • c is 4.18 J g-1 K-1.
  • m is 50 g. Since it's assumed that both solutions have a density of 1 g cm-1, and both solutions have volumes of 25 cm3, this means they both have masses of 25 g. So, mixing these two solutions would give a combined volume of 50 cm3, and so a combined mass of 50 g.

So, we have everything to calculate the temperature change. \Delta T = \frac{2805}{50*4.18} = 13.4 K.
Would you not calculate the mass of water produced in the reaction? I specifically remember being told that the mass is also the mass of water and not the mass of any solids/solutions?
0
reply
Kian Stevens
Badges: 16
Rep:
?
#4
Report 1 year ago
#4
(Original post by Principleturn)
Would you not calculate the mass of water produced in the reaction? I specifically remember being told that the mass is also the mass of water and not the mass of any solids/solutions?
No you wouldn't. The water is simply being formed, not undergoing any reaction; it's the 25 cm3 of ethanoic acid and the 25 cm3 of sodium hydroxide which are reacting and causing an enthalpy change.
What you were very much likely told was that the density of a solution can be assumed to be the density of water. The question wouldn't tell you to assume that the densities of both solutions are equal to that of water for nothing...
0
reply
Principleturn
Badges: 4
Rep:
?
#5
Report Thread starter 1 year ago
#5
(Original post by Kian Stevens)
No you wouldn't. The water is simply being formed, not undergoing any reaction; it's the 25 cm3 of ethanoic acid and the 25 cm3 of sodium hydroxide which are reacting and causing an enthalpy change.
What you were very much likely told was that the density of a solution can be assumed to be the density of water. The question wouldn't tell you to assume that the densities of both solutions are equal to that of water for nothing...
Ok Thanks!
0
reply
looooolddbdsfv
Badges: 3
Rep:
?
#6
Report 1 year ago
#6
i don't understand why you do the enthalpy change multiplied by 0.05, why are we altering the value they have given us? also i dont get why at the end the masses combine to get the answer in cm^3 and why we haven't divided by 100?this is what i would do , i'm assuming its wrong but i don't know why.5.6.1/0.1 x 4.18 = 134.21...any help is appreciated
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Do you have the space and resources you need to succeed in home learning?

Yes I have everything I need (398)
56.29%
I don't have everything I need (309)
43.71%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise