Neutralization reaction
HCL(aq) + NaOH(aq) -> NaCl(s) + H2O(l)
Use your results to determine the netball change of the neutralization reaction, in kJ/mol of NaOH, Write the thermochemical equation for the neutralization reaction.
I'm not sure how to even go about doing this, do I use the molar enthalpies of formation for water and salt? Or?
Enthalpy*
I have two questions.
1. What dos netball mean?
2. What results do you have that are supposed to help you answer this question?
To determine the net enthalpy change (ΔH) of the neutralization reaction in kJ/mol of NaOH, you can use the concept of Thermochemical Equations and the Hess's Law.
Hess's Law states that if a reaction can be expressed as a series of steps, the overall enthalpy change is the sum of the enthalpy changes for each step. In this case, we can use the following steps:
1. The formation of NaOH(s) from its elements:
Na(s) + 1/2 O2(g) + 1/2 H2(g) → NaOH(s) ΔH₁
2. Dissolving solid NaOH in water:
NaOH(s) → NaOH(aq) ΔH₂
3. The reaction of HCl(aq) with NaOH(aq):
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ΔH₃
4. The formation of NaCl(aq) from its elements:
Na(s) + 1/2Cl2(g) → NaCl(aq) ΔH₄
We need to determine the net enthalpy change, ΔH₃, for the neutralization reaction.
Using the concept of Hess's Law, we rearrange and sum up the enthalpy changes for the steps to get:
ΔH₃ = ΔH₁ + ΔH₂ + ΔH₄
Now, to find the values of the enthalpy changes, you can use the molar enthalpies of formation (ΔHf) for the reactants and products involved in each step. The molar enthalpies of formation represent the enthalpy change when one mole of a compound is formed from its elements in their standard states at a given temperature (usually 25°C or 298 K).
Assuming you have the values of ΔHf for NaOH(s), NaOH(aq), HCl(aq), and NaCl(aq), you can substitute these values into the ΔH values for the corresponding steps in the equation above and calculate the net enthalpy change, ΔH₃.
For example, if you have the values in kJ/mol:
ΔH₁ = ΔHf(NaOH(s)) - [ΔHf(Na(s)) + 1/2ΔHf(O2(g)) + 1/2ΔHf(H2(g))]
ΔH₂ = ΔHf(NaOH(aq)) - ΔHf(NaOH(s))
ΔH₄ = ΔHf(NaCl(aq)) - [ΔHf(Na(s)) + 1/2ΔHf(Cl2(g))]
By calculating these values and substituting them back in the equation, you will get the net enthalpy change of the neutralization reaction, ΔH₃, in kJ/mol of NaOH.
The thermochemical equation for the neutralization reaction can be written as:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ΔH₃ (net enthalpy change) kJ/mol NaOH
Remember to use accurate values for ΔHf and to ensure the balancing of the equation.