contestada

a. Use the following data to calculate the enthalpy of hydration for calcium chloride and calcium iodide.
Lattice Energy* ΔHsoln
CaCl_2(s) -2247 kJ/mol -46 kJ/mol
CaI_2(s) -2059 kJ/mol -104 kJ/mol
*Lattice energy is defined as the energy change for the process M^+(g) + X^-(g) -------> MX(s).
b. Based on your answers to part a, which ion, Cl^- or I^-, is more strongly attracted to water?
O Chloride
O Iodide

Respuesta :

mickymike92

Answer:

A. Enthalpy of hydration of CaCl2 = -2293 KJ/mol

Enthalpy of hydration of CaI2 = -2163 KJ/mol

B. Chloride ions

Explanation:

Enthalpy of hydration is the energy required to convert one mole of gaseous ions to aqueousions.

Enthalpy of hydration = Enthalpy of solution - lattice energy

For CaCl2;

Enthalpy of hydration = -46 - 2247 = -2293 KJ/mol

For CaI2;

Enthalpy of hydration = -104 - 2059 = -2163 KJ/mol

B. Since the enthalpy of hydration of CaCl2 is more negative than that of CaI2, it means that chloride ions are more easily attracted to water molecules. This we conclude because the difference in enthalpy values are due to the negative ions as the two compounds share a common positive calcium ion.

ajeigbeibraheem

The enthalpy for hydration of calcium chloride is -2293 kJ/mol. The enthalpy for the hydration of calcium iodide is -2163 kJ/mol. The chloride ion Cl^- is more strongly attracted to water than iodide.

The enthalpy of hydration also the enthalpy of solvation is the quantity of energy released when 1 mole of gaseous ion is diluted.

It can be expressed by using the formula:

[tex]\mathbf{\Delta H_{hyd} = \Delta H_{solution} + \Delta H_{LE}}[/tex]

The enthalpy for hydration of calcium chloride can be expressed as:

[tex]\mathbf{\Delta H_{hyd(cacl_2)} = \Delta H_{solution(caci_2)} + \Delta H_{LE(caci_2)}}[/tex]

[tex]\mathbf{\Delta H_{hyd(cacl_2)} = -46 \ kJ/mol +-2247 \ kJ/mol}[/tex]

[tex]\mathbf{\Delta H_{hyd(cacl_2)} =-2293 \ kJ/mol}[/tex]

The enthalpy for the hydration of calcium iodide can be expressed as:

[tex]\mathbf{\Delta H_{hyd(cal_2)} = \Delta H_{solution(caI_2)} + \Delta H_{LE(caI_2)}}[/tex]

[tex]\mathbf{\Delta H_{hyd(cacl_2)} = -104 \ kJ/mol +-2059 \ kJ/mol}[/tex]

[tex]\mathbf{\Delta H_{hyd(cacl_2)} =-2163 \ kJ/mol}[/tex]

From the above calculation, we can see that the enthalpy of hydration of calcium chloride is greater than that of calcium iodide,

Therefore, we can conclude that the chloride ion Cl^- is more strongly attracted to water than iodide.

Learn more about enthalpy of hydration here:

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