Enthalpy Question - help!

The student carries out a second experiment using 150 cm3 of distilled water instead of
100 cm3 of distilled water. The mass of concentrated sulfuric acid is the same as in the
first experiment.
Predict and explain the effect, if any, of the larger volume of water on the following:
• The temperature change, ΔT
• The calculated value of ΔsolH for H2SO4.

- I understand that the temperature change will be less as the energy will be more spread out in the larger volume of water
- The mark scheme says the that the temperature change would be 7 degrees which I don't get why?
- Also, the enthalpy change of solution doesn't change - could someone explain why please?
Thanks :smile:

Reply 1

TypicalNerd

Original post by kittymityyy

I would need more data to be sure exactly where the final answer comes from. Though in the meantime, it might help if you can try thinking of a useful equation for how to calculate an energy change you know that may be relevant.

Provided you are using the same solvent each time (which in this case they are as distilled water is used), then enthalpy of solution doesn’t change with the volume, amount of substance used etc.

Recall that the magnitude of the enthalpy change = (energy change)/(moles)

This ratio will always be the same, regardless of how much solvent and solute you use.

I think the simplest way to think about it is that every 1 mole of a particular substance stores (or needs to take in, if the substance dissolves endothermically) a fixed quantity of energy and for every gram of this substance you dissolve, you release this fixed amount of energy - independent of how much solvent you are using.

Suppose you have “x” moles of a substance and you dissolve it in water. Suppose each mole of this substance stores Q kJ of energy and so the total amount of energy that can be released upon dissolution is xQ kJ of energy.

The enthalpy change = (energy change)/(moles)
ΔH = (xQ kJ)/(x mol) = Q kJ/mol

(edited 1 year ago)

Reply 2

Maryyyyy5

Original post by kittymityyy

The student carries out a second experiment using 150 cm3 of distilled water instead of
100 cm3 of distilled water. The mass of concentrated sulfuric acid is the same as in the
first experiment.
Predict and explain the effect, if any, of the larger volume of water on the following:
• The temperature change, ΔT
• The calculated value of ΔsolH for H2SO4.
- I understand that the temperature change will be less as the energy will be more spread out in the larger volume of water
- The mark scheme says the that the temperature change would be 7 degrees which I don't get why?
- Also, the enthalpy change of solution doesn't change - could someone explain why please?
Thanks :smile:

To get from 150 to 100 you divide by 1.5, so they take the temp change which is 10.5 and divide it by 1.5 too to get 7

Reply 3

Nitrotoluene

Original post by kittymityyy

The student carries out a second experiment using 150 cm3 of distilled water instead of
100 cm3 of distilled water. The mass of concentrated sulfuric acid is the same as in the
first experiment.
Predict and explain the effect, if any, of the larger volume of water on the following:
• The temperature change, ΔT
• The calculated value of ΔsolH for H2SO4.
- I understand that the temperature change will be less as the energy will be more spread out in the larger volume of water
- The mark scheme says the that the temperature change would be 7 degrees which I don't get why?
- Also, the enthalpy change of solution doesn't change - could someone explain why please?
Thanks :smile:

To better understand the dynamics of the experiment, let us look at the effect of increasing the volume of water. Specifically, we will see how this change affects the change in temperature, ΔT, and the calculated value of ΔsolH for H2SO4.Using 150 cm3 of distilled water instead of 100 cm^3 changes the thermal behaviour of the experiment.
The heat released during the reaction between H2O and H2SO4 remains constant, but a larger volume of water distributes this heat over a larger mass of water.
Consequently, the temperature change, ΔT, must be less than the temperature change in the experiment with 100 cm^3 of water.
This is because heat dissipates at a slower rate, causing a more gradual rise in the temperature of the solution until it reaches its final value.
Furthermore, the calculated value of ΔsolH for H2SO4 varies by the increased volume of water.
While the mass of concentrated sulphuric acid remains unchanged, increasing the volume of water leads to an overall increase in the total mass of the system.
This, in turn, affects the calculation of ΔsolH because a larger volume of water works as a heat sink for the solution. Therefore, the calculated value of ΔsolH will be lower than the value in the original experiment with 100 cm^3 of water.