A student gets the following data for the heat of neutralization trials using 1.

Question

A student gets the following data for the heat of neutralization trials using 1.0 M NaOH and 1.0 M HCl.

A. Mass of calorimeter /g trial1:43.5

B. Mass HCl and calorimeter / g trial1:64.0

C. Mass calorimeter, HCl and NaOH / g trial1:85.0

D1. Mass HCl / g (B - A) trial1:.......

D2. Mass NaOH /g (C - B) trial1:.......

Mass of Solution (D1 + D2) trial1:.......

Initial temp. HCl measured / C (therm 1) trial1:22.1

Initial temp. HCl corrected / C trial1:........

Initial temp. NaOH measured / C (therm 2) trial1:21.6

Initial temp. NaOH corrected / C trial1:.......

E. Average Initial Temp corrected / C trial1:.......

Final temp. measured / C (therm 1) trial1:28.8

F. Final temp. corrected / C trial1:.......

G. Temp. change corrected / C o (F - E) trial1:........

Using this data, determine the Molar Enthalpy of Neutralization for this trial. The mcalib calib is 0.969 and b is 2.0 C for both Thermometer 1 and Thermometer 2 and B the calculated heat capacity of the calorimeter was 33 J °C-1 .

Explanation / Answer

A. Mass of calorimeter /g trial1: 43.5

B. Mass HCl and calorimeter / g    trial1: 64.0

C. Mass calorimeter, HCl and NaOH / g  trial1: 85.0

D1. Mass HCl / g (B - A)                           trial1: 29.0

D2. Mass NaOH /g (C - B)                        trial1: 21.0

Mass of Solution (D1 + D2)                       trial1: 50.0

Initial temp. HCl measured / C (therm 1) trial1: 22.1

Initial temp. HCl corrected / C                        trial1: 22.1 + 0.969 = 23.069

Initial temp. NaOH measured / C (therm 2) trial1: 21.6

Initial temp. NaOH corrected / C                    trial1: 21.6 + 2 = 23.6

E. Average Initial Temp corrected / C            trial1: 23.3345

Final temp. measured / C (therm 1)       trial1: 28.8

F. Final temp. corrected / C                           trial1: 28.8 + 0.969 = 29.769

G. Temp. change corrected / C o (F - E)      trial1: 29.769 - 23.3345 = 6.4345

Heat of neutralization : (msol * Cpwater * deltaT) + (Cpcal * deltaT)

H = (50g * 4.18 J/gºC * 6.4345) + (33 J/ºC * 6.4345) = 1557.15 J/mol = 1.56 kJ

We have 21g of NaOH, which are aproximately, taking density of 1 kg/L and 1 M, 0.021 moles

We have 29 g of HCl, which are 0.029 moles

We divide the heat by 0.021 moles to get molar heat of neutralization:

1.56 kJ / 0.021 moles = 74.2857 kJ/mol

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