Question

Lab Notebook Trial 1 Trial 2 Trial 3 100 ml 100 ml 23 °C 100 mL 24 C 38 c 278 mol 25 mol Volume of HCI Initial temperature of HCI Volume of NaOH Initial temperature of NaOH FInal solution temperature Moles of H' Moles of OH Limiting reagent Heat transfer for solution 100 mL 24 C 100 mL 24 C 40℃ 278 mol 25 mol 22℃ 100 ml 23 C 38℃ 278 mol 25 mol OH OH OH HC kJ kJ Enthalpy of neutralization per mole of limiting reagent k/mol K/mol Mean enthalov of neutralization per mole of limiting reagent Redo Take

Answer 1

The specific heat of water is 4.184 J/g0C. The density of both HCL and NaOH is 1.00 g/mL

Assume that specific heats HCL and NaOH are also 4.1984 J/g C.

For Trial 1

To calculate the heat actually evolved, use the formula

                                            q = m c Δt

To convert ml to grams, multiply volume with density (m = V X D)

(100 mL + 100 mL) = 200 mL of solution

100 mL X 1 g/ml = 2 00 grams of solution.

Find the temperature change.

Δt = t final – t initial = 40 ⁰C – 24 ⁰C =16⁰C

q = m c Δt

q = 200 grams X 4.184   J /g⁰C x 16 ⁰C

q= 13388.8 J

q = 13.39KJ

This is the heat gained by the water, but in fact it is the heat lost by the reacting HCl and NaOH, therefore q = - 13.39KJ    

Moles of HCl = 0.278

Moles of NaOH = 0.25   

NaOH is limiting reagent   

Enthalpy of neutralization /mol of limiting reagent = -13.39KJ /0.25     

   =- 52.56 KJ/mol  

For Trial 2

Δt = t final – t initial = 38 0C – 23.5 0C =14.50C

q = m c Δt

q = 200 grams X 4.184   J /g0C x 14.5 0C

q= -12133.6 J

q = - 12.13KJ      

Moles of HCl = 0.278

Moles of NaOH = 0.25                 

Enthalpy of neutralization /mol of limiting reagent = -12.13KJ /0.25     

     =- 48.52 KJ/mol  

For Trial 3

Δt = t final – t initial = 38 0C – 22.5 0C =15.5 0C

q = m c Δt

q = 200 grams X 4.184   J /g0C x 15.5 0C

q= 12970.4 J

q = -12.97KJ   

Enthalpy of neutralization /mol of limiting reagent =- 12.97KJ /0.25     

     = -51.88 KJ/mol  

Mean enthalpy of neutralization /mol of limiting reagent

= (-52.56 KJ/mol   + -48.52 KJ/mol   +-51.88 KJ/mol ) /3

Mean enthalpy of neutralization /mol of limiting reagent - 50.99 KJ/mol