enthalpy change

50.0 cm3 of 1.00 mol dm−3 NaOH is neutralised by 50.0 cm3 of 1.00 mol dm−3 HNO3. The temperature increases by 6.0 °C. The experiment is repeated using: 25.0 cm3 of 1.00 mol dm−3 NaOH and 25.0 cm3 of 1.00 mol dm−3 HNO3. What is the increase in temperature in the second experiment?

A 1.5°C
B 3.0°C
C 6.0°C
D 12.0°C

Reply 1

temanbaik

Original post by pengot983

50.0 cm3 of 1.00 mol dm−3 naoh is neutralised by 50.0 cm3 of 1.00 mol dm−3 hno3. The temperature increases by 6.0 °c. The experiment is repeated using: 25.0 cm3 of 1.00 mol dm−3 naoh and 25.0 cm3 of 1.00 mol dm−3 hno3. What is the increase in temperature in the second experiment?

A 1.5°c
b 3.0°c
c 6.0°c
d 12.0°c

option d

Reply 2

scimus63

not sure but I would say...

for 50ml of 1M acid and alkali the number of moles reacting is 0.05moles of each. This gives a temp rise of 6 degrees
so
for 25ml of 1M acid and alkali the number of moles reacting is just 0.025 moles of each, so since half the number of moles and also half the volume of water produced or has to be heated so would the temp rise still be 6 degrees??????

Reply 3

NCHEM

Original post by scimus63

Yes you are correct

Reply 4

cg12456

Original post by scimus63

not sure but I would say...
for 50ml of 1M acid and alkali the number of moles reacting is 0.05moles of each. This gives a temp rise of 6 degrees
so
for 25ml of 1M acid and alkali the number of moles reacting is just 0.025 moles of each, so since half the number of moles and also half the volume of water produced or has to be heated so would the temp rise still be 6 degrees??????

Can you please explain

Reply 5

puprceee

Original post by cg12456

Can you please explain

wanted to know the explanation as well

Reply 6

UtterlyUseless69

Original post by puprceee

wanted to know the explanation as well

In the first experiment, double the amounts of the same chemicals are used. Because the concentrations of the chemicals are the same, it suggests that the following statements must be true:

-Experiment 2 produces half as many moles of water as experiment 1.

-The total mass of the solution in experiment 2 is half as much as it is in experiment 1.

-The specific heat capacities of the solutions in experiments 1 and 2 are the same.

Let’s now recall that ΔH = -mcΔθ/n

Where ΔH is the enthalpy change, m is the mass of the solution, c is the specific heat capacity of the solution, Δθ is the temperature change and n is the number of moles used.

Because the same reaction is taking place in both experiments, ΔH must have the same value in both experiments.

We can rearrange the above equation to make ΔH the subject:

Δθ = -(nΔH)/(mc)

Let’s say for argument’s sake that in experiment 1, the number of moles of water produced is 0.05 mol, the mass of the solution is 100 g, c is 4.18 J/g/K and ΔH is x J/mol. Plugging these into the equation:

Δθ = -(0.05x)/(100 * 4.18) = -0.00012x (2 sf)

In experiment 2, ΔH and c have the same value, but both the number of moles of water and the mass of the solution are halved (i.e n = 0.025 and m = 50 g). Thus, we have:

Δθ = -(0.025x)/(50 * 4.18) = -0.00012x (2 sf)

Notice that these are the same expression? This means that the temperature change has to be the same for both experiments and so in this case, the correct answer is 6°C.

(edited 1 week ago)

Reply 7

WordsFiddle

Instead of calculating, since you won't have time for it during your exam, you can figure out relations.
n x ΔH = mc ΔT
we know that ΔH and c are the same.
so n ∝ mΔT
let n be the number of moles for first reaction. Let m be the mass in first reaction. Let ΔT be the temperature change for first reaction. Let Δt be the temperature change for second reaction.
Then for second reaction, moles = 0.5n, mass= 0.5m.
(ΔT/Δt) = (n/0.5n) x (0.5m/m) (or ÷ m/0.5m) which gives us Δt= ΔT.
Just a useful trick to know for mcq papers.