Question

50.0 mL of 2.00 mol/L HNO3 solution and 50.0 mL of 1.00 mol/L NaOH solution, both at 20.0 degree Celsius, were mixed in a calorimeter. Calculate the molar heat of neutralization of HNO3 in kJ/mol if:

(1) final temperature was 28.9 degree Celsius;

(2) the mass of the overall solution was 102.0 g;

(3) the heat capacity of the calorimeter was 25.0 J/C;

(4) and assume that the specific heat of solution is the same as water, 4.184 J/(g C);

Answer 1

Moles of NaOH taken = M x V = 1.00 mol/L x 0.050 L = 0.050 mol

Moles of HNO3 taken = M x V = 2.00 mol/L x 0.050 L = 0.100 mol

The balanced chemical reaction for the neutralization reaction is

HNO3(aq) + NaOH(aq) ------- > H2O(l) + NaNO3(aq)

1 mol, ------- 1 mol

1 mol HNO3 reacts with 1 mol NaOH.

=> 0.050 mol NaOH that will react with the moles of HNO3 = 0.050 mol NaOH x ( 1mol HNO3 / 1 mol NaOH)

= 0.050 mol HNO3

Hence HNO3 is in excess and only 0.050 mol of HNO3 is neutralized.

Increase in temperature, T = 28.9 - 20.0 = 8.9 DegC

Total heat released during the neutralization reaction, Qt = Heat absorbed by solution + Heat absorbed by calorimeter

=> Qt = mxSxT + CxT

=> Qt = (102.0 g x 4.184 J/gxC x 8.9 C) + (25.0 J/C x 8.9 C)

=> Qt = 3798.235 J + 222.5 J

=> Qt = 4020.735 J

Hence 4020.735 J heat is released when 0.050 mol HNO3 is neutralized.

=> Amount of heat that will release when 1 mol HNO3 is neutralized = (4020.735 J / 0.050 mol) x 1 mol = 80415 J/mol

=  80415 J/mol x (1 kJ / 1000 J)

= 80.4 kJ / mol (answer)