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Chemistry enthalphy change question

In a past paper this is the question:

In a different experiment 50.0 cm3 of 0.500 mol dm–3 aqueous hydrochloric acid are
reacted with 50.0 cm3 of 0.500 mol dm–3 aqueous sodium hydroxide.
NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) ΔH = –57.1 kJ mol–1
The initial temperature of each solution is 18.5 °C
Calculate the maximum final temperature of the reaction mixture.
Assume that the specific heat capacity of the reaction mixture, c = 4.18 J K–1 g–1
Assume that the density of the reaction mixture = 1.00 g cm–3

The markscheme has this as the answer:

n(HCl) or n(NaOH) = 50 x 0.500 / 1000 = 0.025 moles
q = –ΔH x n = 57.1 x 0.025 = 1.4275 kJ
ΔT = q/mc
ΔT = (1.4275 x 1000) / (100 x 4.18) = 3.4(2) °C
Final Temperature = 18.5 + 3.4 = 21.9 °C

Which I get up until
ΔT = (1.4275 x 1000) / (100 x 4.18) = 3.4(2) °C
... why do they times by 1000? Does the volume need to be in dm^3 or something?
q = 1.4275 is in kilojoules
for the ΔT = q/mc you want q to be in J
Original post by bl0bf1sh
q = 1.4275 is in kilojoules
for the ΔT = q/mc you want q to be in J


wow I feel dumb, I guess the exam stress is getting to me

thank you and sorry you had to witness that
Original post by KookieCrumble
wow I feel dumb, I guess the exam stress is getting to me

thank you and sorry you had to witness that

no worries, it happens to everyone!
good luck in your exams :smile:
Reply 4
why do we use the mole of only one substance and not both?
Original post by bl0bf1sh
q = 1.4275 is in kilojoules
for the ΔT = q/mc you want q to be in J

Why did they multiply 4.18 by 100
Original post by Mathswarrior
Why did they multiply 4.18 by 100

ΔT = q ÷ mc = (1.4275 x 1000) ÷ (100 x 4.18)
4.18 J K–1 g–1 is the specific heat capacity (c) of the reaction mixture
100 g is the mass (m) of the mixture (they are aqueous solutions. The mass of 1 cm3 of water is 1 g, so assume that 50 cm3 + 50 cm3 (the two solutions) is about 100 g) :smile:
Original post by bl0bf1sh
ΔT = q ÷ mc = (1.4275 x 1000) ÷ (100 x 4.18)
4.18 J K–1 g–1 is the specific heat capacity (c) of the reaction mixture
100 g is the mass (m) of the mixture (they are aqueous solutions. The mass of 1 cm3 of water is 1 g, so assume that 50 cm3 + 50 cm3 (the two solutions) is about 100 g) :smile:

Ohh, thanks so much. I got confused but it's good now.
why is q= negative of delta H ????????????
Reply 10
Original post by Ajsjdhdhdh
why is q= negative of delta H ????????????


It's just showing that it's an exothermic reaction but when you use the delta H value in your calculation you don't use the negative
Original post by bl0bf1sh
q = 1.4275 is in kilojoules
for the ΔT = q/mc you want q to be in J

why are we multiplying 4.18 by 100 then?
Original post by mahii786
why are we multiplying 4.18 by 100 then?

100 g is the mass, which you multiply by the specific heat capacity
Reply 13
Original post by bl0bf1sh
ΔT = q ÷ mc = (1.4275 x 1000) ÷ (100 x 4.18)
4.18 J K–1 g–1 is the specific heat capacity (c) of the reaction mixture
100 g is the mass (m) of the mixture (they are aqueous solutions. The mass of 1 cm3 of water is 1 g, so assume that 50 cm3 + 50 cm3 (the two solutions) is about 100 g) :smile:

how are we getting 50cm3 and 50cm3, do u mean like 1g=100cm?
Reply 14
Original post by mahii786
how are we getting 50cm3 and 50cm3, do u mean like 1g=100cm?

The total volume of the mixture is 50 cm3 (of HCl) + 50 cm3 (of NaOH) = 100 cm3
1 cm3 of water has a mass of 1 g (by definition), so 100 cm3 of water would have a mass of 100 g
Since the HCl and NaOH are aqueous solutions, we can assume that they’re going to be mostly water and therefore will have approximately the same mass as pure water :smile:
Reply 15
Original post by bl0bf1sh
The total volume of the mixture is 50 cm3 (of HCl) + 50 cm3 (of NaOH) = 100 cm3
1 cm3 of water has a mass of 1 g (by definition), so 100 cm3 of water would have a mass of 100 g
Since the HCl and NaOH are aqueous solutions, we can assume that they’re going to be mostly water and therefore will have approximately the same mass as pure water :smile:

ohh I understand now thank you!!
Reply 16
why is moles 0.025 not 0.05😞
Original post by qmmxrq
why is moles 0.025 not 0.05😞

Hello qmmxrq!

The reason why the molar amounts are 0.025 and not 0.050 is because the reaction uses 25.0 millimoles of both NaOH and HCl, not 50.0 millimoles. The reaction is 1:1 stoichiometric, so the number of moles of NaOH and HCl is the same.

Number of moles of HCl or NaOH = n = concentration (c) x volume (V)
where (c) is the concentration in moles per litre (mol/L) and (V) is the volume in litres (L).
(c) = 0.500 mol dm^-3 = 0.500 mol/L (since 1 dm^3 = 1 L)
n = 0.500 mol/L x 0.05 L (because 50 cm^3 = 0.05 L) = 0.025 mol = 25.0 mmol (millimoles)
So the number of moles in the 50 cm^3 or 0.05 L solution of HCl or NaOH is 0.025 mol = 25.0 mmol (millimoles)
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Sandro
(edited 4 months ago)
Original post by KookieCrumble
In a past paper this is the question:
In a different experiment 50.0 cm3 of 0.500 mol dm–3 aqueous hydrochloric acid are
reacted with 50.0 cm3 of 0.500 mol dm–3 aqueous sodium hydroxide.
NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) ΔH = –57.1 kJ mol–1
The initial temperature of each solution is 18.5 °C
Calculate the maximum final temperature of the reaction mixture.
Assume that the specific heat capacity of the reaction mixture, c = 4.18 J K–1 g–1
Assume that the density of the reaction mixture = 1.00 g cm–3
The markscheme has this as the answer:
n(HCl) or n(NaOH) = 50 x 0.500 / 1000 = 0.025 moles
q = –ΔH x n = 57.1 x 0.025 = 1.4275 kJ
ΔT = q/mc
ΔT = (1.4275 x 1000) / (100 x 4.18) = 3.4(2) °C
Final Temperature = 18.5 + 3.4 = 21.9 °C
Which I get up until
ΔT = (1.4275 x 1000) / (100 x 4.18) = 3.4(2) °C
... why do they times by 1000? Does the volume need to be in dm^3 or something?

They divide by 1000 as the KJ needs to be in J

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