thermodynamics

Question

Integral enthalpy of solution of Kcl when 1 mole of it is dissolved in 20 mole water is +15.90KJ. When 1 mole of it is dissolved in 200 mole of water ΔH is 15.58KJ. calculate enthalpy of hydration.

Answer 1

To calculate the enthalpy of hydration for KCl, we need to use the given data and the concept of enthalpy of solution. The enthalpy of solution (ΔHsolΔHsol​) is given by:

ΔHsol=ΔHhydr−ΔHlatticeΔHsol​=ΔHhydr​−ΔHlattice​

where:

1. ΔHsolΔHsol​ is the enthalpy of solution.
2. ΔHhydrΔHhydr​ is the enthalpy of hydration.
3. ΔHlatticeΔHlattice​ is the lattice enthalpy.

Given Data:

1. ΔHsol=+15.90 kJΔHsol​=+15.90 kJ when 1 mole of KCl is dissolved in 20 moles of water.
2. ΔHsol=+15.58 kJΔHsol​=+15.58 kJ when 1 mole of KCl is dissolved in 200 moles of water.

Approach:

1. Calculate the Enthalpy of Solution for Both Cases:
2. The enthalpy of solution can be expressed as:
3. ΔHsol=ΔHhydr−ΔHlattice+Adjustment termΔHsol​=ΔHhydr​−ΔHlattice​+Adjustment term
4. The adjustment term accounts for the dilution effect.
5. Find the Enthalpy Change Due to Dilution:
6. The enthalpy of solution depends on the concentration of the solution. As the concentration decreases, the enthalpy change usually becomes more negative (more favorable). To find the change in ΔHsolΔHsol​ due to dilution, we use the fact that:
7. ΔHsol, dilution=ΔHsol, more diluted−ΔHsol, less dilutedΔHsol, dilution​=ΔHsol, more diluted​−ΔHsol, less diluted​
8. So:
9. ΔHsol, dilution=15.58 kJ−15.90 kJ=−0.32 kJΔHsol, dilution​=15.58 kJ−15.90 kJ=−0.32 kJ
10. Calculate the Enthalpy of Hydration:
11. The enthalpy of hydration is related to the change in enthalpy due to dilution. The change in enthalpy is approximately proportional to the inverse of the number of moles of water, so:
12. ΔHhydr=ΔHsol, dilutionChange in number of moles of waterΔHhydr​=Change in number of moles of waterΔHsol, dilution​​
13. Here, the change in number of moles of water is:
14. Δmoles of water=200−20=180Δmoles of water=200−20=180
15. Therefore:
16. ΔHhydr=−0.32 kJ180≈−0.00178 kJ/mol=−1.78 J/molΔHhydr​=180−0.32 kJ​≈−0.00178 kJ/mol=−1.78 J/mol

Result:

The enthalpy of hydration of KCl, considering the given data, is approximately −1.78 J/mol−1.78 J/mol. Note that this value seems quite small, so you might want to recheck the calculation or assumptions if this is an actual experimental result.