Answer to Question #215471 in General Chemistry for mani

Question

200 cm³ of 1.0M HCL and 200 cm³ of 1.0M NaOH at 33.1^oC were mixed in a simple calorimeter. The temperature of the calorimeter assembly increased to 39.3^oC. Calculate the enthalpy of neutralization in kJ/mol. The heat capacity of the calorimeter is 101.8 J/K. Assume the specific heat of the solution to be 4.186 J/g.K and density of solution to be 1.0 g/cm³.

Answer 1

Total Volume of the Solution =200mL +200mL=400mL

Total Mass of Solution =Volume(mL)×Density (g/mL)

Total Mass of Solution =400mL×1g/mL

=400g

Heat released during the reaction (q) =$mc\Delta T$

q=400g×4.186J/g/K×(39.3-33.1)$\degree C$

q=10381.28J =10.38128kJ

Heat absorbed by the calorimeter =101.8J/K×(39.3-33.1)=631.16J =0.63116kJ

HCl=1M×0.2L=0.1mol

NaOH=1M×0.2L=0.1mol

For neutralization therefore=(0.1mol+0.1mol)=0.2mol

$\therefore\Delta H$ =10.38128kJ+0.63116kJ$\frac{10.38128kJ+0.63116kJ}{0.2mol}$ =55.06kJ/mol

Hence enthalpy of reaction is 55.06kJ/mol