Calculate the enthalpy change to be expected for NaOH(s) to NaOH(aq)?
1/2H2(g) + 1/2 Cl2(g) -- HCl(g) -92.3 kJ/mol
Na(s) + 1/2 O2(g) + 1/2H2(g) -- NaOH(s) -426.8kJ/mol
Na(s) + 1/2Cl2(g) -- NaCl(s) -411.1kJ/mol
H2(g) + 1/2O2(g) -- H2o(l) -285.8kJ/mol
HCl(g) -- HCl(aq) -75.2kJ/mol
NaCl(s) -- NaCl(aq) 4.0kJ/mol
been working at this problem for days, and can't seem to get it to balance. Any help is appreciated.
Answer:
You are missing one equation: HCl (aq) + NaOH (aq) --> NaCl (aq) + H2O (l), -56.0 kJ.
NaOH (s) --> Na (s) + 1/2 H2 (g) + 1/2 O2 (g) +426.8 kJ/mol
NaCl (aq) + H2O (l) --> NaOH (aq) + HCl (aq) +56.0 kJ/mol
Na (s) + 1/2 Cl2 (g) --> NaCl (s) -411.1 kJ/mol
NaCl (s) --> NaCl (aq) +4.0 kJ/mol
HCl (g) --> 1/2 H2 (g) + 1/2 Cl2 (g) +92.3 kJ/mol
HCl (aq) --> HCl (g) +75.2 kJ/mol
H2 (g) + 1/2 O2 (g) --> H2O (l) -285.8 kJ/mol
When all these equations are added together, the only thing left is NaOH (s) --> NaOH (aq), thus adding all the enthalpy values is the enthalpy change for that reaction; -42.6 kJ/mol.
You can confirm this using the standard enthalpy of formation values of NaOH (s) and NaOH (aq), which are -426.8 (given in your problem) and -469.6 for NaOH (aq), so NaOH (s) --> NaOH (aq), ∆H = -469.6 - (-426.8) = -42.8 (close to what I got from using Hess' Law).
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