Answer:
51.54°C the final temperature of the calorimeter contents.
Explanation:
[tex]HCl+NaOH\rightarrow H_2O+NaCl,\Delta H=-56 kJ/mol[/tex]
[tex]moles=Molarity\times Volume (L)[/tex]
Molarity of HCl= 0.50 M
Volume of HCl= 150.0 mL = 0.150 L
Moles of HCl= n
[tex]n=0.50 M\times 0.150 L=0.075 mol[/tex]
Molarity of NaOH= 1.00 M
Volume of NaOH= 50.0 mL = 0.050 L
Moles of NaOH= n'
[tex]n'=1.00 M\times 0.050 L=0.050 mol[/tex]
Since moles of NaOH are less than than moles of HCl. so energy release will be for neutralization of 0.050 moles of naOH by 0.050 moles of HCl.
n = 0.050
[tex]-56 kJ/mol=-\frac{Q}{n}[/tex]
[tex]Q=56 kJ/mol\times 0.050 kJ/mol=2.8 kJ=2800 J[/tex]
(1 kJ= 1000 J)
The energy change released during the reaction = 2800 J
Volume of solution = 150.0 mL + 50.0 mL = 200.0 mL
Density of the solution (water) = 1.00g/mL
Mass of the solution , m= 200 mL × 1.00 g/mL = 200 g
Now , calculate the final temperature by the solution from :
[tex]q=mc\times (T_{final}-T_{initial})[/tex]
where,
q = heat gained = 2800 J
c = specific heat of solution = [tex]4.184 J/^oC[/tex]
[tex]T_{final}[/tex] = final temperature = [tex]?[/tex]
[tex]T_{initial}[/tex] = initial temperature = [tex]48.2^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]2800 J=200.0 g\times 4.184 J/^oC\times (T_{final}-48.2)^oC[/tex]
[tex]T_{final}= 51.54^oC[/tex]
51.54°C the final temperature of the calorimeter contents.
