find formation enthalphy of Mg(OH)2 with below details
2Mg+o2----->2MgO=-1203.6
Mg(OH)2---->MgO+H2O =+37.1
2H2+O2---->2H2O =-571.7
To find the formation enthalpy of Mg(OH)2, which is the enthalpy change when one mole of Mg(OH)2 is formed from its constituent elements in their standard states, we can use the Hess's law.
Hess's law states that if a chemical reaction can be expressed as the sum of two or more reactions, then the enthalpy change of the overall reaction is the sum of the enthalpy changes of the individual reactions.
In this case, we can use the following reactions to determine the enthalpy change for the formation of Mg(OH)2:
1. 2Mg + O2 → 2MgO ΔH = -1203.6 kJ/mol (Given)
2. Mg(OH)2 → MgO + H2O ΔH = +37.1 kJ/mol (Given)
3. 2H2 + O2 → 2H2O ΔH = -571.7 kJ/mol (Given)
Now, let's manipulate these reactions to find the formation reaction of Mg(OH)2:
1. We need to reverse the second reaction to get MgO and H2O as reactants and Mg(OH)2 as the product:
MgO + H2O → Mg(OH)2 ΔH' = -37.1 kJ/mol (Reverse the sign of ΔH)
2. We need to multiply the second reaction by 2 to cancel out the MgO, since the first reaction has 2MgO as a product:
2(MgO + H2O → Mg(OH)2) ΔH" = 2(-37.1) = -74.2 kJ/mol (Multiply ΔH' by 2)
3. Now we can add the manipulated reactions together:
2Mg + O2 + 2H2 → 2H2O + Mg(OH)2 ΔHf = -1203.6 - 571.7 - 74.2 = -1849.5 kJ/mol
Therefore, the formation enthalpy of Mg(OH)2 is -1849.5 kJ/mol.