Question

TC01E03 25.0 mL of 1.00 M NaOH solution is mixed with 25.0 mL of 1.05 M HCl in an open calorimeter. You can assume volumes of these solutions are additive. The density of the resulting solution is 1.03 g/mL and its heat capacity is 3.90 J/K/g. The temperature of the calorimeter and its contents rose by 6.70 °C. What is the molar enthalpy of reaction for NaOH reacting with HCI? You can neglect the heat capacity of the calorimeter. Select one: a. -2247 kJ/mol O b.-53.8 kJ/mol O C. -1.35 kJ/mol O d. - 1124 kJ/mol e. -52.3 kJ/mol

Answer 1

Volume of NaOH = 25 ml  

Concentration of NaOH = 1.00 M

Moles of NaOH = 25 x 1 / 1000 = 0.025 Moles

Volume of HCl = 25 ml

Concentration of HCl = 1.05 M

Moles of HCl = 25 x 1.05 / 1000 = 0.02625 Moles

Moles of NaCl produced = 0.025 Moles

Using the following formula we can calculate the hjeat of the reaction

Q = mc∆T

Q = heat energy (Joules, J), m = mass of a substance (g)

c = specific heat (units J/g∙^oC), ∆ is a symbol meaning "the change in"

∆T = change in temperature (^oC Celcius)

m = 25 + 25 = 50 ml / 1.03 g/ml = 51.5 gm

c = 3.9 J/g∙^oC  ∆T = 6.7 ^oC

Q = 51.5 gm x 3.9 J/g∙^oC x 6.7 ^oC = 1345.695 Joules or 1.345 Kilo Joules

Enthalpy of the reaction = 1.345 Kilo Joules / 0.025 Moles = 53.82 kJ /mol

Hence option B is correct answer.