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 4 months ago)