Respuesta :
Answer:
Heat of the reaction per mole of NaOH = 46.02 kJ/mol
Explanation:
The reaction between HCl (strong acid) and NaOH(strong base) is a neutralization reaction which yields a salt NaCl and water
[tex]HCl + NaOH \rightarrow NaCl + H2O[/tex]
The heat (q) of a reaction is given as:
[tex]q = m*c*\Delta T=m*c*(T2-T1)-------(1)[/tex]
where m = mass of the system
c = specific heat
T1 and T2 are the initial and final temperatures
It is given that:
Volume of HCl = 500.0 ml
Volume of NaOH = 500.0 ml
Density of HCl and NaOH = 1.000 g/ml
[tex]Mass = Density*Volume[/tex]
[tex]Mass(HCl) = Mass(NaOH) = 1.000g/ml*500.0ml = 500.0 g[/tex]
Total mass of the solutions, m = 500.0 +500.0 = 1000.0 g
c = 4.184 J/g/c
T1 = 25.6 C
T2 = 26.70 C
Substituting appropriate values in equation (1) gives:
[tex]q = 1000.0 g*4.184 J/gC *(26.7-25.6)C = 4602.4 J=4.602kJ[/tex]
Now, the number of moles of NaOH is:
[tex]Moles(NaOH)=Molarity(NaOH)*Volume(NaOH)= 0.200moles/L*0.500L = 0.100moles[/tex]
Heat of reaction/mole NaOH is:
[tex]=\frac{q}{moles(NaOH)}=\frac{4.602kJ}{0.100moles}=46.02kJ/mol[/tex]
