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?
To find the heat of solution of anhydrous CaCl2 in a large volume of water, we need to calculate the enthalpy change for the reaction:
CaCl2(s) + xH2O(l) -> CaCl2 * xH2O(s)
First, let's calculate the molar enthalpy change for the formation of the hydrated CaCl2, which is given as -97.0 kJ/mol. We can assume that this value is for the reaction with 6 moles of water (x = 6).
Now, let's calculate the molar enthalpy change for dissolving solid CaCl2 in water:
ΔH = ΔH hydration + ΔH solution
ΔH hydration = -97.0 kJ/mol (given)
ΔH solution = ΔH hydration - Σ(n * ΔH dissolving)
ΔH dissolving = 14.6 kJ/mol (given)
Since there is 1 mole of CaCl2 reacting in the formation of the hydrated compound, we can substitute n = 1:
ΔH solution = -97.0 kJ/mol - (1 * 14.6 kJ/mol)
= -97.0 kJ/mol - 14.6 kJ/mol
= -111.6 kJ/mol
Therefore, the heat of solution of anhydrous CaCl2 in a large volume of water is -111.6 kJ/mol.
To find the heat of solution of anhydrous CaCl2 in water, we need to find the heat change when one mole of anhydrous CaCl2 dissolves in water.
From the given information, we know that when one mole of CaCl2(s) reacts with 6 moles of water to form CaCl2 * 6H2O(s), the enthalpy change (delta H) is -97.0 kJ.
The equation for the dissolution of anhydrous CaCl2 in water can be written as:
CaCl2(s) -> Ca2+(aq) + 2 Cl-(aq)
Since the dissolution process does not involve water molecules as reactants, we are not directly given the heat of solution of anhydrous CaCl2 in water. However, we can use Hess's Law to find the heat of solution.
Hess's Law states that the total enthalpy change for a reaction is independent of the pathway taken, as long as the initial and final conditions are the same. Therefore, we can use the enthalpy change for the reaction:
CaCl2(s) + 6H2O(l) -> CaCl2 * 6H2O(s)
and the enthalpy change for the reaction:
CaCl2 * 6H2O(s) -> Ca2+(aq) + 2 Cl-(aq)
to determine the heat of solution of anhydrous CaCl2.
Since the reaction CaCl2(s) + 6H2O(l) -> CaCl2 * 6H2O(s) has a delta H of -97.0 kJ, we can reverse the reaction and change the sign of delta H:
CaCl2 * 6H2O(s) -> CaCl2(s) + 6H2O(l) (delta H = +97.0 kJ)
Next, we need to find the enthalpy change for the reaction:
CaCl2(s) -> Ca2+(aq) + 2 Cl-(aq)
To determine this, we can sum up the enthalpies of the reverse reactions of the formation of the reactants:
Ca2+(aq) + 2 Cl-(aq) -> CaCl2(s) (delta H1)
6 H2O(l) -> 6H2O(l) (delta H2 = 0, as it is the same substance)
Therefore, the enthalpy change for the reaction:
CaCl2(s) -> Ca2+(aq) + 2 Cl-(aq)
is equal to the enthalpy change of the reverse of the formation reaction (CaCl2(s) + 6H2O(l) -> CaCl2 * 6H2O(s)) minus the enthalpy change of the formation of water molecules:
delta H = delta H1 - delta H2
Substituting the values:
delta H = 97.0 kJ - 0 kJ
delta H = 97.0 kJ
Therefore, the heat of solution of anhydrous CaCl2 in a large volume of water is 97.0 kJ/mol.