Question

The aluminum cup inside your calorimeter weighs 41.55 g. You add 59.21 g of 1.0 M acetic acid solution and 50.03 g of 1.0 M sodium hydroxide solution to the calorimeter. Both solutions have an initial temperature of 19.9 °C, and the final temperature after addition is 26.8 °c. What is the molar enthalpy of neutralization, in units of kJ/mol? Assume that: the calorimeter is completely insulated the heat capacity of the empty calorimeter is the heat capacity of the aluminum cup: 0.903 J g1 °C-1 the density of the two solutions is the same as that of water: 1.00 g/mL the heat capacity of the two solutions is the same as that of water: 4.184 J g~1 Dorfor all. eulationswiths

Answer 1

As the calorimeter is completely insulated, heat released during the neutralisation reaction of the acetic acid and sodium hydroxide is absorbed by the solution of the two and the aluminium cup.

Mass of solution=Mass of acetic acid+Mass of sodium hydroxide

=59.21 g+50.03 g=109.24 g

Rise in temperature of the solution=Rise in temperature of the Aluminium cup=Final temperature of solution-initial temperature of the solution

=26.8°C-19.9°C=6.9°C

Heat absorbed by the solution and the calorimeter

=Mass of solution x specific heat of solution x rise in temperature of solution + Mass of aluminium cup x specific heat of aluminium cup x rise in temperature

=109.24 gx4.184 J/g°C x6.9°C+41.55 g x 0.903 J/g°Cx6.9°C

=3153.71 J+258.89 J

=3412.60 J

Heat released in the neutralisation reaction of acetic acid and sodium hydroxide=-heat absorbed by the solution

=-(3412.60 J)

=-3412.60 J/1000 J/kJ=3.412 kJ (1 kJ=1000 J)

Molar mass of acetic acid (CH₃COOH)=2xMolar mass of C+4xMolar mass of H+2xMolar mass of O=2x12 g/mol+4x1 g/mol+2x16 g/mol=24 g/mol+4 g/mol+32 g/mol=60 g/mol

Number of moles of acetic acid=Given mass/Molar mass

=59.21 g/60 g/mol=0.99 mol

Molar mass of sodium hydroxide (NaOH)=Molar mass of Na+Molar mass of O+Molar mass of H=23 g/mol+16 g/mol+1 g/mol=40 g/mol

Number of moles of sodium hydroxide=Given mass/molar mass=50.03 g/40 g/mol=1.25 mol

The balanced chemical equation for reaction between acetic acid and sodium hydroxide is

NaOH + CH3COOH + CH3COONa + H2O

As per the balanced chemical reaction

1 mol acetic acid reacts with 1 mol NaOH

So 0.99 mol acetic acid reacts with 0.99 mol NaOH

But we have 1.25 mol NaOH which is more than the required amount

So acetic acid is the limiting reagent

So molar enthalpy of neutralisation=heat released in neutralisation/number of moles of acetic acid

=-3.412 kJ/0.99 mol=-3.45 kJ/mol