Using the following thermochemical data, what is the change in enthalpy for the following reaction?
Ca(OH)¿(aq) + HCl(aq) d CaCl¿(aq) + H¿O(l)
CaO(s) + 2HCl(aq) d CaCl¿(aq) + H¿O(l), DH = -186 kJ
CaO(s) + H¿O(l) d Ca(OH)¿(s), DH = -65.1 kJ
To find the change in enthalpy (ΔH) for the given reaction, we can use Hess's Law, which states that if a reaction can be expressed as the sum of two or more reactions, then the enthalpy change for the overall reaction is the sum of the enthalpy changes of the individual reactions.
In this case, we can see that the desired reaction can be written as the sum of the two given reactions:
1) CaO(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) with ΔH = -186 kJ
2) CaO(s) + H2O(l) → Ca(OH)2(s) with ΔH = -65.1 kJ
We need to adjust these reactions so that Ca(OH)2 is on the reactant side and Ca(OH) is on the product side, like in the desired reaction. To do this, we can reverse reaction 2 and multiply it by 2 to balance the moles of Ca(OH)2:
3) 2Ca(OH)2(s) → 2CaO(s) + 2H2O(l) with ΔH = 2 × (-65.1 kJ) = -130.2 kJ
Now, we can add reactions 1 and 3 to get the desired reaction:
4) (2CaO(s) + 2HCl(aq)) + (2Ca(OH)2(s) → (CaCl2(aq) + H2O(l)) + (2CaO(s) + 2H2O(l))
Simplifying this equation: 2CaO(s) + 2HCl(aq) + 2Ca(OH)2(s) → CaCl2(aq) + H2O(l) + 2CaO(s) + 2H2O(l)
Cancelling out common terms: 2HCl(aq) + 2Ca(OH)2(s) → CaCl2(aq) + H2O(l) + 2H2O(l)
Simplifying further: 2HCl(aq) + 2Ca(OH)2(s) → CaCl2(aq) + 3H2O(l)
The ΔH for the desired reaction (Ca(OH)2(aq) + HCl(aq) → CaCl2(aq) + H2O(l)) can be obtained by adding the ΔH values of reactions 1 and 3:
ΔH = ΔH1 + ΔH3 = -186 kJ + (-130.2 kJ) = -316.2 kJ
Therefore, the change in enthalpy for the given reaction is -316.2 kJ.