standard enthalpies of formation
H20(l) = -286 LiOH(s) = -487
standard enthalpies of solution for LiOH(s) to split into it's ions is 21
all values are in KJmol-1
Use the data given to calculate a value for the enthalpy change for the reaction of lithium with water.
I get how to do the cycle and I've seen the mark scheme and I don't understand why the 21 becomes negative.
I can draw my cycles. I use the equation in hess's law:
AH1 = AH2+AH3
= 286 - 487
= -201 and that is the wrong answer, it has -21 aswell so the answer is -222. I would have thought that if the solution was being formed then it would have a positive value as you are going in the right direction, is the mark scheme wrong or is there something I don't understand.
Thank you for any help you can provide.
The enthalpy of solution for LiOH(s) is the enthalpy change for the reaction of lithium with water. The enthalpy of solution is the energy required to break the ionic bonds in the solid LiOH(s) and form the ions Li+ and OH- in solution. Since energy is required to break the bonds, the enthalpy of solution is a positive value. Therefore, the enthalpy change for the reaction of lithium with water is -222 kJ/mol.
First, let's break down the reaction of lithium with water into steps:
Step 1: Formation of LiOH(s) from Li(s)
Li(s) → LiOH(s) (1)
Step 2: Dissolution of LiOH(s) in water to form Li+ and OH- ions
LiOH(s) → Li+(aq) + OH-(aq) (2)
Step 3: Formation of H2O(l) from H+(aq) and OH-(aq)
H+(aq) + OH-(aq) → H2O(l) (3)
Now, let's use the given enthalpies of formation and enthalpy of solution to calculate the enthalpy change for the overall reaction.
Step 1: The standard enthalpy change for the formation of LiOH(s) from Li(s) is given as -487 kJ/mol.
Step 2: The standard enthalpy change for the dissolution of LiOH(s) into Li+ and OH- ions is given as +21 kJ/mol. Note that the positive sign indicates that the process is endothermic (energy is absorbed).
Step 3: The standard enthalpy change for the formation of H2O(l) from H+(aq) and OH-(aq) is given as -286 kJ/mol.
To calculate the enthalpy change for the overall reaction, follow the steps:
1. Add the enthalpies for Steps 1 and 2 since they are in the forward direction:
AH1 = AH1 + AH2
= -487 kJ/mol + 21 kJ/mol
= -466 kJ/mol
2. Add the enthalpy change for Step 3, but since it is in the reverse direction, you need to reverse the sign:
AH3 = -(-286 kJ/mol) (note the negative sign)
= +286 kJ/mol
3. Sum them up to get the overall enthalpy change:
Overall: AHLi(s) + H2O(l) = AH1 + AH2 + AH3
= -466 kJ/mol + 286 kJ/mol
= -180 kJ/mol
So, the enthalpy change for the reaction of lithium with water (formation of LiOH(s) and H2O(l)) is -180 kJ/mol.
It seems there was an error in your calculation. The correct answer should be -180 kJ/mol, not -222 kJ/mol. The positive value for the enthalpy change of solution for LiOH(s) simply indicates that the process is endothermic.
To calculate the enthalpy change for the reaction of lithium with water, we can use Hess's law and the given data. The enthalpy change of the reaction can be expressed as:
ΔH = ΔH1 + ΔH2 + ΔH3
where ΔH1 is the standard enthalpy of formation of water (H2O(l)), ΔH2 is the standard enthalpy of formation of lithium hydroxide (LiOH(s)), and ΔH3 is the standard enthalpy of solution for LiOH(s).
Given the following values:
ΔH1 = -286 kJ/mol (negative because it is the enthalpy of formation of a compound)
ΔH2 = -487 kJ/mol (negative because it is the enthalpy of formation of a compound)
ΔH3 = 21 kJ/mol (positive because it is the enthalpy of solution)
By substituting the values into the equation, we have:
ΔH = -286 + (-487) + 21
= -286 - 487 + 21
= -752 + 21
= -731 kJ/mol
So, the enthalpy change for the reaction of lithium with water is -731 kJ/mol.
Regarding your question about the positive value for the enthalpy of solution (ΔH3): It is important to note that the enthalpy of solution represents the energy change associated with dissolving one mole of a substance into its ions. In this case, the positive value of 21 kJ/mol indicates that it requires energy to break the ionic bonds in lithium hydroxide and convert it into its ions. Therefore, as this process requires energy, it contributes to the overall endothermicity of the reaction.