Dissolving solid CaCl2 H2O in a large volume of water is endothermic to the extent of 14.6 kj/mol. Also given: CaCl2(s)+6H2O(l)->CaCl2* 6H2O(s) delta H = -97.0 kj what is the heat of solution of anhydrous CaCl2 in a large volume of water?
Is that a typo first line. Perhaps should be CaCl2.6H2O. If so just add the two equations to obtain
CaCl2(s) ==> CaCl2(aq) about -82 kJ/mol but that isn't exact.
To find the heat of solution of anhydrous CaCl2 in a large volume of water, we need to consider the heat of hydration of CaCl2(s) and the heat of dissolution of anhydrous CaCl2.
The given equation for the dissolution of hydrated CaCl2 is:
CaCl2(s) + 6H2O(l) -> CaCl2 * 6H2O(s) ΔH = -97.0 kJ
This means that for every mole of hydrated CaCl2 that dissolves, 97.0 kJ of heat is released.
Now, let's consider the process of dissolving anhydrous CaCl2:
CaCl2(s) -> CaCl2(aq) ΔH = ?
Since anhydrous CaCl2 does not contain water molecules, we need to consider the heat of hydration of CaCl2(s) to form CaCl2 * 6H2O(s). From the given information, we know that the heat of hydration is -97.0 kJ/mol.
To calculate the heat of solution of anhydrous CaCl2, we need to subtract the heat of hydration from the heat of dissolution of hydrated CaCl2:
Heat of Solution = Heat of Dissolution - Heat of Hydration
Heat of Solution = -97.0 kJ - (-97.0 kJ)
Heat of Solution = -97.0 kJ + 97.0 kJ
Heat of Solution = 0 kJ/mol
Therefore, the heat of solution of anhydrous CaCl2 in a large volume of water is 0 kJ/mol.
To find the heat of solution of anhydrous CaCl2 in a large volume of water, we need to use the given information:
1. The heat of hydration of CaCl2·6H2O (CaCl2 with 6 water molecules) is -97.0 kJ/mol. This means that when one mole of CaCl2·6H2O dissolves in water, it releases 97.0 kJ of heat.
2. The heat of solution of solid CaCl2·H2O (CaCl2 with 1 water molecule) is given as 14.6 kJ/mol. This means that when one mole of solid CaCl2·H2O dissolves in water, it absorbs 14.6 kJ of heat.
To find the heat of solution of anhydrous CaCl2, we need to subtract the heat of hydration of CaCl2 from the heat of solution of CaCl2·H2O.
Since the stoichiometric ratio of anhydrous CaCl2 to CaCl2·6H2O is 1:1, we can assume that the heat of solution of anhydrous CaCl2 would be the same as the heat of solution of CaCl2·H2O. Therefore, the heat of solution of anhydrous CaCl2 in water would also be 14.6 kJ/mol.
In conclusion, the heat of solution of anhydrous CaCl2 in a large volume of water is 14.6 kJ/mol.