Use the following data to calculate the heat of hydration, ΔHhydr, for the ions (K+ and BrO3−) in potassium bromate (KBrO3). Also write a chemical equation describing this hydration process. ΔHlattice = -745 kJ mol−1 ; ΔHsoln = 41.1 kJ mol−1
The chemical equation for the hydration of K+ and BrO3- in KBrO3 can be written as:
KBrO3(s) + aq → K+(aq) + BrO3-(aq)
To determine the heat of hydration, we need to consider the energy changes involved in breaking up the crystal lattice and dissolving the compound in water, as well as the energy released when the ions are hydrated.
ΔHsoln = ΔHlattice + ΔHhydr
where ΔHsoln is the enthalpy change of solution, ΔHlattice is the enthalpy change of lattice energy, and ΔHhydr is the enthalpy change of hydration.
Rearranging the equation above, we can calculate ΔHhydr as:
ΔHhydr = ΔHsoln - ΔHlattice
ΔHhydr = 41.1 kJ mol-1 - (-745 kJ mol-1)
ΔHhydr = 786.1 kJ mol-1
Therefore, the heat of hydration for K+ and BrO3- ions in KBrO3 is 786.1 kJ mol-1.
Chemical equation:
KBrO3(s) + aq → K+(aq) + BrO3-(aq) + ΔHhydr
To calculate the heat of hydration, ΔHhydr, for the ions K+ and BrO3− in potassium bromate (KBrO3), you can use the equation:
ΔHhydr = ΔHsoln - ΔHlattice
Given:
ΔHlattice = -745 kJ mol−1
ΔHsoln = 41.1 kJ mol−1
Substituting these values into the equation:
ΔHhydr = 41.1 kJ mol−1 - (-745 kJ mol−1)
= 41.1 kJ mol−1 + 745 kJ mol−1
= 786.1 kJ mol−1
Therefore, the heat of hydration, ΔHhydr, for the ions K+ and BrO3− in potassium bromate (KBrO3) is 786.1 kJ mol−1.
The chemical equation describing this hydration process is:
KBrO3(s) + aq → K+(aq) + BrO3−(aq)
where (s) represents the solid KBrO3 and (aq) represents the aqueous ions K+ and BrO3−.