Question

9) 50.0 mL of 0.250 M sodium hydroxide at 19.5 oC was added to 50.0 mL of 0.250M hydrochloric acid also at 19.5 oC in a calorimeter. After mixing the ` solution temperature rises to 21.21oC. What is the heat of reaction?

Answer 1

Ans :

Volume of solution = 50.0 mL + 50.0 mL = 100 mL

density of solution = 1.04 g/mL

so mass of solution = 1.04 g/mL x 100 mL = 104 g

specific heat of solution = 4.017 J/g^oC

each mol NaOH reacts with 1 mol HCl to form 1 mol each salt and water

so mol water = 0.250 M x 0.050 L = 0.0125 mol

heat = mass x specific heat x change in temperature

putting values :

= 104 g x 4.017 J/g^oC x (21.21^oC - 19.5^oC)

= -714.4 J

heat is evolved in the reaction so a negative sign

Heat of reaction = heat / mol water

= -714.4 J / 0.0125 mol

= -57150 J/mol

= -57.1 KJ/mol