Calculate the standared enthalpy of formation of solid Mg(OH)2 given the following data:
2Mg(s) + O29(g) →2MgO(s) ∆H = -1203.6kJ
Mg(OH)2(s) →MgO(s) + H2O(l) ∆H= +37.1kJ
2H2(g) +O2(g) →2H2O(l) ∆H = -571.7kJ
Use eqn 1 as is.
Reverse eqn 2 and multiply by 2. (also multiply delta H by 2 and reverse the sign).
Use eqn 3 as is.
You should get
2Mg + 2O2 + 2H2 ==> 2Mg(OH)2 which is just twice what you want. So add the delta Hs from above and take half of it.
Standard enthalpy of formation is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states. In this case, we need to calculate the standard enthalpy of formation of Mg(OH)2.
First, let's analyze the given equations and data:
2Mg(s) + O2(g) → 2MgO(s) ΔH = -1203.6 kJ (Equation 1)
Mg(OH)2(s) → MgO(s) + H2O(l) ΔH = +37.1 kJ (Equation 2)
2H2(g) + O2(g) → 2H2O(l) ΔH = -571.7 kJ (Equation 3)
We need to manipulate these equations to find the desired enthalpy. Taking Equation 1 and reversing it gives us:
2MgO(s) → 2Mg(s) + O2(g) ΔH = +1203.6 kJ (Equation 4)
Next, we multiply Equation 2 by 2 to match the number of moles of MgO:
2Mg(OH)2(s) → 2MgO(s) + 2H2O(l) ΔH = +2 * 37.1 kJ = +74.2 kJ (Equation 5)
Next, we multiply Equation 3 by 2 to match the number of moles of H2O:
4H2(g) + 2O2(g) → 4H2O(l) ΔH = 4 * (-571.7 kJ) = -2286.8 kJ (Equation 6)
Adding Equation 4, Equation 5, and Equation 6 together gives us the balanced equation:
2Mg(OH)2(s) → 2Mg(s) + 2H2O(l) ΔH = 1203.6 kJ + 74.2 kJ - 2286.8 kJ = -1009 kJ
Therefore, the standard enthalpy of formation of solid Mg(OH)2 is approximately -1009 kJ. To make it more "entertaining," let's call it the "Hot Mess enthalpy" because this reaction seems quite messy energetically!
To calculate the standard enthalpy of formation of solid Mg(OH)2, we can use Hess's Law. Hess's Law states that the enthalpy change of a reaction is independent of the pathway taken.
First, let's write down the balanced chemical equations for the given reactions:
1. 2Mg(s) + O2(g) → 2MgO(s) (∆H = -1203.6 kJ)
2. Mg(OH)2(s) → MgO(s) + H2O(l) (∆H = +37.1 kJ)
3. 2H2(g) + O2(g) → 2H2O(l) (∆H = -571.7 kJ)
We want to find the enthalpy of formation of Mg(OH)2, which means we need to combine the given reactions in a way that the desired compound is formed.
If we look at reaction 2, we see that MgO is a product. In reaction 1, MgO is also a product. So, we can reverse reaction 1 and use it to cancel out MgO as a product in reaction 2.
Reversed reaction 1: 2MgO(s) → 2Mg(s) + O2(g) (∆H = +1203.6 kJ)
Now, let's add reaction 2 and the reversed reaction 1 together:
Mg(OH)2(s) + 2MgO(s) → 2Mg(s) + O2(g) + H2O(l)
To cancel out MgO on both sides, we need to multiply reaction 2 by 2 and reverse reaction 1:
2Mg(OH)2(s) + 4MgO(s) → 4Mg(s) + 2O2(g) + 2H2O(l) (∆H = 2 * +37.1 kJ + +1203.6 kJ)
However, we need to cancel out O2 on both sides as well. For this, we can use reaction 3. Multiply reaction 3 by 2:
4Mg(OH)2(s) + 4MgO(s) + 4H2O(l) → 4Mg(s) + 4H2O(l) (∆H = 4 * +37.1 kJ + 2 * -571.7 kJ + 2 * +1203.6 kJ)
Now, observe that there are 4 moles of H2O on both sides, so they cancel out:
4Mg(OH)2(s) + 4MgO(s) → 4Mg(s) (∆H = 4 * +37.1 kJ + 2 * -571.7 kJ + 2 * +1203.6 kJ)
Finally, simplify the equation:
Mg(OH)2(s) → Mg(s) (∆H = +1609.6 kJ)
Therefore, the standard enthalpy of formation of solid Mg(OH)2 is +1609.6 kJ.
To calculate the standard enthalpy of formation of solid Mg(OH)2, we need to use the given data and apply Hess's Law. Hess's Law states that the overall enthalpy change of a reaction is independent of the pathway taken.
The standard enthalpy of formation (ΔHf°) is the change in enthalpy when 1 mole of a compound is formed from its constituent elements in their standard states.
We can break down the desired reaction into a series of known reactions:
1. Reverse the second reaction equation:
MgO(s) + H2O(l) -> Mg(OH)2(s) ΔH = -37.1 kJ (change sign because reverse reaction)
2. Multiply the second reaction equation by 2 to match the coefficients of MgO in the first equation:
2MgO(s) + 2H2O(l) -> 2Mg(OH)2(s) ΔH = -2 * -37.1 kJ = 74.2 kJ
3. Reverse the first reaction equation:
2MgO(s) -> 2Mg(s) + O2(g) ΔH = -(-1203.6 kJ) = 1203.6 kJ (change sign because reverse reaction)
4. Divide the first reaction equation by 2 to match the coefficient of MgO in the second equation:
MgO(s) -> Mg(s) + 0.5O2(g) ΔH = 1203.6 kJ / 2 = 601.8 kJ
5. Multiply the third reaction equation by 2 to match the coefficient of O2 in the fourth equation:
2Mg(s) + O2(g) -> 2MgO(s) ΔH = 2 * -601.8 kJ = -1203.6 kJ
Now we can add the equations together to cancel out the intermediates:
2Mg(s) + O2(g) -> 2MgO(s) ΔH = -1203.6 kJ
2MgO(s) + 2H2O(l) -> 2Mg(OH)2(s) ΔH = 74.2 kJ
----------------------------------------------
2Mg(s) + O2(g) + 2H2O(l) -> 2Mg(OH)2(s) ΔH = -1203.6 kJ + 74.2 kJ = -1129.4 kJ
The standard enthalpy of formation of solid Mg(OH)2 can be calculated as follows:
ΔHf°(Mg(OH)2) = ΔH of the overall reaction / moles of Mg(OH)2 formed
= -1129.4 kJ / 2
= -564.7 kJ/mol
Therefore, the standard enthalpy of formation of solid Mg(OH)2 is -564.7 kJ/mol.