Respuesta :
Answer:
ΔH(neu) = 30.68 kJ/mol
Explanation:
The reaction between HCl (strong acid) and NaOH(strong base) is a neutralization reaction which yields sodium chloride 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 = 1000 ml
Volume of NaOH = 100.0 ml
Density of HCl and NaOH = 1.000 g/ml
[tex]Mass = Density*Volume[/tex]
[tex]Mass(NaOH) = 1.000g/ml*1000 ml = 1000 g[/tex]
[tex]Mass(HCl) = 1.000g/ml*100.0 ml = 100.0 g[/tex]
Total mass of the solutions, m = 1000 +100.0 = 1100.0 g
c = 4.184 J/g/c
T1 = 20.0C
T2 = 22.0 C
Substituting appropriate values in equation (1) gives:
[tex]q = 1100.0 g*4.184 J/gC *(22.0-20.0)C = 9204.8 J=9.205kJ[/tex]
Now, the number of moles of NaOH is:
[tex]Moles(NaOH)=Molarity(NaOH)*Volume(NaOH)= 0.300moles/L*1.000L = 0.300moles[/tex]
Enthalpy of neutralization is:
[tex]=\frac{q}{moles(NaOH)}=\frac{9.205kJ}{0.300moles}=30.68kJ/mol[/tex]
